Lecture Notes in Mathematics
Arkansas Tech University
Department of Mathematics
Introductory Notes in Ordinary
Differential Equations for Physical
Sciences and Engineering
Marcel B. Finan
c
All
Rights Reserved
May 4, 2017
Preface
Differential equations arise from the study of problems in virtually every area
of physical sciences as well as in engineering. Many mathematical models of
real world phenomena lead to a differential equation problem for which we
seek its solution. In this course, we will explore analytical, numerical and
graphical techniques for finding the solutions.
To gain a solid grasp of differential equations, it is essential for the reader to
solve ALL the exercises listed at the end of each section. Answers to most
exercises are available at the end of the book.
A solution guide is available by request: Email [email protected]
Marcel B. Finan
Russellville, Arkansas
May 2011
i
ii
PREFACE
Contents
Preface
i
1 Basic Terminology
3
2 Existence and Uniqueness of Solutions to First Order Linear
IVP
13
3 Analytical Solution: The Method of Integrating Factor
19
4 Existence and Uniqueness of Solutions to First Order Nonlinear IVP
29
5 Separable Differential Equations
35
6 Exact Differential Equations
43
7 Substitution Techniques: Bernoulli and Riccati Equations
51
8 Graphical Solution: Direction Field of y 0 = f (t, y)
59
9 Numerical Solutions to ODEs: Euler’s Method and its Variants
69
10 Second Order Linear Differential Equations: Existence and
Uniqueness Results
75
11 The General Solution of 2nd Order Linear Homogeneous Equations
81
1
2
CONTENTS
12 Second Order Linear Homogeneous Equations with Constant
Coefficients: Distinct Characterisitc Roots
91
13 Characteristic Equations with Repeated Roots
97
14 Characteristic Equations with Complex Roots
103
15 Series Solutions of Differential Equations
111
16 The Structure of the General Solution of Linear Nonhomogeneous Equations
117
17 The Method of Variation of Parameters
123
18 The Laplace Transform: Basic Definitions and Results
129
19 Further Studies of Laplace Transform
141
20 The Laplace Transform and the Method of Partial Fractions151
21 Laplace Transforms of Periodic Functions
159
22 Convolution Integrals
167
Cheat Sheets
173
Answer Key
181
Index
223
1 Basic Terminology
In many models, we will have equations involving the derivatives of a dependent variable y with respect to one or more independent variables and are
interested in discovering this function y. Such equations are referred to as
differential equations (abbreviated DE). They arise in many applications
such as population growth, decay of radioactive substance, the motion of a
falling object, electrical network, Newton’s law of cooling and many more
models.
A First Source of Differential Equations: Vertical Motion of an
Object
We consider the model of uniform motion. Suppose that an object initially
at height y0 is moving straight up or down with initial velocity v0 . Let y(t)
denote the distance of the object from the ground , v(t) the object’s velocity,
and a(t) the object’s acceleration at time t. We assume y to be positive in
the upward direction.
If air resistance is neglected, then by Newton’s second law, which states
that the net force is equal to the product of mass and acceleration, we have
ma(t) = −mg. The negative sign on the right-hand of the equation is due
to the fact that acceleration due to gravity is pointing downward. Using the
fact that a(t) = y 00 (t) and eliminating the mass, we obtain the equation
y 00 = −g.
To find the velocity v(t) we integrate for a first time and obtain
v(t) = y 0 (t) = −gt + C1 .
Since the initial velocity is v0 , we obtain C1 = v(0) = v0 so that
v(t) = −gt + v0 .
3
4
1 BASIC TERMINOLOGY
Integrating for the second time we find the position function
1
y(t) = − gt2 + v0 t + C2 .
2
Since y0 is the initial height, we find C2 = y0 and so
1
y(t) = − gt2 + v0 t + y0 .
2
Example 1.1
An object is dropped from the top of a cliff that is 144 feet above ground
level.
(a) When will the object reach ground level?
(b) What is the velocity with which the object strikes the ground?
Solution.
(a) The motion of the object translates to the differential equation y 00 = −32
with solution y(t) = −16t2 + 144. The object reaches
ground level when
q
= 3 sec. The object
y(t) = 0 or 16t2 = 144. Solving for t we find t = 144
16
will reach the ground 3 seconds after it is dropped from the cliff.
(b) The object strikes the ground with velocity v(3) = −32(3) = −96 ft/sec
Basic Terms of Differential Equations
We next discuss some basic notions of differential equations. There are two
types of differential equations: ordinary and partial differential equations.
By an ordinary differential equation (abbreviated ODE) we mean an
equation that involves an unknown function (the dependent variable) of
a single variable, its independent variable, and one or more of its derivatives. The highest order derivative that appears in the equation is known
as the order of the equation. Thus, an nth order ordinary differential
equation is an equation of the form
y (n) = f (t, y, y 0 , · · · , y (n−1) )
or alternatively
G(t, y, y 0 , y 00 , · · · , y n ) = 0.
A first-order ordinary differential equation, for example, takes the form
f (t, y(t), y 0 (t)) = 0, and may alternatively be written as
y 0 (t) = g(t, y(t))
5
for all t in the interval of existence of y.
Similarly, a second-order ordinary differential equation takes the form
f (t, y(t), y 0 (t), y 00 (t)) = 0 or y 00 = h(t, y, y 0 ).
Example 1.2
Determine the order of each equation.
2
(a) y 0 + 2ty = e−t .
2
+ 6y(t) = 0.
(b) ddt2y − 5 dy
dt
00
0
(c) y + 3ty + 2y = sin (5t).
Solution.
(a) This is a first order differential equation because the highest derivative is
the first derivative.
(b) and (c) are second order differential equations since the highest derivative
in each equation is the second order derivative
When a dependent function is a function of two or more independent variables
then the derivatives are known as partial derivatives. An equation that
involves a function of more than two independent variables and its partial
derivatives is called a partial differential equation (abbreviated PDE).
For example, the wave equation is a partial differential equation of the
form
1 ∂ 2u
∂ 2u
−
= 0.
∂x2 c2 ∂t2
In this course, when we use the term differential equation, we’ll mean an
ordinary differential equation.
A solution of a differential equation is a function that satisfies the equation:
When you substitute this function and its derivatives into the differential
equation, you get a true mathematical statement.
Example 1.3
Show that the function y = 100 + e−t is a solution to the differential equation
y 0 = 100 − y.
Solution.
Indeed, finding the first order derivative of y we have y 0 = −e−t . Also, 100 −
y = 100 − (100 + e−t ) = −e−t . Thus, y 0 = 100 − y so that y = 100 + e−t is a
solution to the given DE
6
1 BASIC TERMINOLOGY
Example 1.4 (A piecewise-defined solution)
Consider the differential equation ty 0 − 4y = 0 on the interval (−∞, ∞).
Verify that the piecewise-defined function
−t4 , t < 0
y=
t4 , t ≥ 0
is a solution.
Solution.
For t < 0, we have ty 0 − 4y = t(−t4 )0 − 4(−t4 ) = −4t4 + 4t4 = 0. For t ≥ 0,
we have ty 0 − 4y = t(t4 )0 − 4t4 = 4t4 − 4t4 = 0. Thus, the given function is a
solution
Solving a differential equation means finding all possible solutions of the
equation.
Example 1.5
Solve the differential equation:
y 00 = −2t.
Solution.
Integrating twice, all the solutions have the form
t3
+ C1 t + C2
3
Note that the function of the previous example defines all the solutions to
the differential equation. Such a function will be referred to as the general
solution. The constants C1 and C2 are called the parameters. Specific
values of C1 and C2 determine what is called a particular solution. To find
a particular solution additional conditions on the values of the function or
its derivatives must be given. Such conditions are called initial conditions.
A differential equation together with a set of initial conditions is called an
initial value problem (abbreviated IVP).
y(t) = −
Example 1.6
Consider the differential equation y 00 (t) − 1 = 0.
(a) Find the general solution of this equation.
(b) Find the solution that satisfies the initial conditions y(1) = 1 and y 0 (1) =
4.
7
Solution.
(a) Integrating twice we find the general solution
y(t) =
t2
+ C1 t + C2 .
2
(b) Since y 0 (t) = t + C1 and y 0 (1) = 4, we find 4 = 1 + C1 so that C1 = 3.
2
Hence, y(t) = t2 + 3t + C2 . Now, since y(1) = 1, we have 1 = 21 + 3 + C2 .
Solving for C2 we find C2 = − 52 . Hence, the solution to the IVP
y 00 (t) − 1 = 0
y 0 (1) = 4, y(1) = 1
is
y(t) =
t2
5
+ 3t −
2
2
The graph of a particular solution is called a solution curve. The function
y(t) = Ce−3t + 2t + 1 is the general solution to the differential equation
y 0 + 3y = 6t + 5. A family of solution curves is shown in Figure 1.1. Notice
for C 6= 0 the solution curves have an oblique asymptote with equation
y(t) = 2t + 1.
Figure 1.1
Sometimes a differential equation possesses a solution that cannot be obtained by assigning values to the parameters in a family of solutions. Such a
solution is called a singular solution.
8
1 BASIC TERMINOLOGY
Example 1.7
Using the method of separation of variables (Chapter 5), the non-zero solu1
2
tions to the differential equation y 0 = ty 2 are given by y(t) = ( t4 + C)2 . Find
the singular solution.
Solution.
The function y(t) ≡ 0 is a solution to the differential equation. This is a
singular solution since it cannot be obtained from the family for any choice
of the parameter C. The general solution consists of all the solutions of the
2
form y(t) = ( t4 + C)2 together with the zero solution
9
Practice Problems
Problem 1.1
A car starts from rest and accelerates in a straight line at 1.6 m/sec2 for 10
seconds.
(a) What is its final speed?
(b) How far has it travelled in this time?
Problem 1.2
An object is thrown upward at time t = 0. After t seconds, its height is
y(t) = −4.9t2 + 7t + 1.5
meters above the ground.
(a) From what height was the object thrown?
(b) What is the initial velocity of the object?
(c) What is the acceleration due to gravity?
Problem 1.3
An object thrown in the air on a planet in a distant galaxy is at height
y(t) = −25t2 + 72t + 40 feet at time t seconds after it is thrown. What is
the acceleration due to gravity on that planet? With what velocity was the
object thrown? From what height?
Problem 1.4
An object is dropped from a 400-foot tower. When does it hit the ground
and how fast is it going at the time of the impact?
Problem 1.5
A ball that is dropped from a window hits the ground in five seconds. How
high is the window? (Give your answer in feet.)
Problem 1.6
A ball is thrown straight up from ground level and reaches its greatest height
after 5 seconds. Find the initial velocity of the ball and the value of its
maximum height above ground level.
Problem 1.7
At time t = 0 an object having mass m is released from rest at a height y0
above the ground. Let g represent the constant gravitational acceleration.
Derive an expression for the impact time (the time at which the object strikes
the ground). What is the velocity with which the object strikes the ground?
10
1 BASIC TERMINOLOGY
Problem 1.8
At time t = 0, an object of mass m is released from rest at a height of
252 ft above the floor of an experimental chamber in which gravitational
acceleration has been slightly modified. Assume (instead of the usual value
ft/sec2 , where of 32 ft/sec2 ), that the acceleration has the form 32− sin πt
4
is a positive constant. In addition, assume that the object strikes the ground
exactly 4 sec after release. Can this information be used to determine the
constant ? If so, determine .
Problem 1.9
Find the order of the following differential equations.
(a) ty 00 + y = t3 .
(b) y 0 + y 2 = 2.
(c) sin (y 000 ) + 3t2 y = 6t.
Problem 1.10
What is the order of the differential equation?
(a) y 0 (t) − 1 = 0.
(b) y 00 (t) − 1 = 0.
(c) y 00 (t) − 2ty(t) = 0.
1
t
= 0.
(d) y 00 (t)(y 0 (t)) 2 − y(t)
Problem 1.11
In the equation
∂u ∂u
−
= x − 2y,
∂x ∂y
identify the independent variable(s) and the dependent variable.
Problem 1.12
Classify the following equations as either ordinary or partial.
2
(a) (y 000 )4 + (y0 )t2 +4 = 0.
(b)
∂u
∂x
+ y ∂u
=
∂y
y−x
.
y+x
(c) y 00 − 4y = 0.
Problem 1.13
Solve the equation y 000 (t) − 2 = 0 by computing successive antiderivatives.
11
Problem 1.14
Solve the initial-value problem
dy
= 3y, y(0) = 50.
dt
What is the domain of the solution?
Problem 1.15
For what real value(s) of λ is y = cos λt a solution of the equation y 00 +9y = 0?
Problem 1.16
For what value(s) of m is y = emt a solution of the equation y 00 +3y 0 +2y = 0?
Problem 1.17
Show that y(t) = et is a solution to the differential equation
2
2
00
0
y − 2+
y + 1+
y = 0.
t
t
Problem 1.18
Show that any function of the form y(t) = C1 cos ωt + C2 sin ωt satisfies the
differential equation
d2 y
+ ω 2 y = 0.
2
dt
Problem 1.19
Suppose y(t) = 2e−4t is the solution to the initial value problem y 0 + ky =
0, y(0) = y0 . Find the values of k and y0 .
Problem 1.20
Consider t > 0. For what value(s) of the constant n, if any, is y(t) = tn a
solution to the differential equation
t2 y 00 − 2ty 0 + 2y = 0?
Problem 1.21
(a) Show that y(t) = C1 e2t + C2 e−2t is a solution of the differential equation
y 00 − 4y = 0, where C1 and C2 are arbitrary constants.
(b) Find the solution satisfying y(0) = 2 and y 0 (0) = 0.
(c) Find the solution satisfying y(0) = 2 and limt→∞ y(t) = 0.
12
1 BASIC TERMINOLOGY
Problem 1.22
Suppose that the graph below is the particular solution to the initial value
problem y 0 (t) = m + 1, y(1) = y0 . Determine the constants m and y0 and
then find the formula for y(t).
Problem 1.23
Suppose that the graph below is the particular solution to the initial value
problem y 0 (t) = mt, y(t0 ) = −1. Determine the constants m and t0 and then
find the formula for y(t).
Problem 1.24
Show that y(t) = e2t is not a solution to the differential equation y 00 + 4y = 0.
Problem 1.25
Consider the initial-value problem
y 0 + 3y = 6t + 5, y(0) = 3.
(a) Show that y = Ce−3t + 2t + 1 is a solution to the above differential
equation.
(b) Find the value of C.
2 Existence and Uniqueness of
Solutions to First Order Linear
IVP
Solutions to differential equations can be given in one of√the following forms:
• by an explicit formula: For example, the function y = √t3 + 1 is an explicit
solution to the initial value problem 2yy 0 = 3t2 , y(1) = 2;
ty
is
• by an implicit equation: The solution y to the equation y 0 = − 1+ye
1+tety
ty
defined implicitly by the equation t + y + e = C;
• by a power series representation. For example the general solution to the
equation (t2 − 4)y 00 + 3ty 0 + y = 0 is given by
3 4
5 6
1 3
1 5
1 7
1 2
t +
t + · · · +C2 t + t + t +
t + ··· ;
y(t) = C1 1 + t +
8
128
1024
6
30
140
• numerically (i.e., Euler’s methods);
• graphically (i.e., direction fields).
Before worrying about how to solve an initial value problem, either analytically, graphically, or numerically, one should first resolve the basic issues
of existence and uniqueness. First, does a solution exist? If, not, it makes no
sense trying to find one. Second, is the solution uniquely determined by the
initial conditions? Otherwise, the differential equation probably has little
relevance for physical applications since we cannot use it as a predictive tool.
In this section we discuss the conditions needed to guarantee the existence of
a unique solution to first order linear initial value problems. We start with
the definition of a first order linear differential equation.
Any differential equation that can be written in the form
y 0 + p(t)y = g(t)
13
(2.1)
142 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP
where p(t) and g(t) are functions with common domain a < t < b, is called a
first order linear differential equation. The term linear is used because
L(y) = y 0 + p(t)y is linear in y. That is, L(αy1 + βy2 ) = αL(y1 ) + βL(y2 ). A
DE that is not linear is called non-linear.
In mathematics and physics, linear generally means “simple” and non-linear
means “complicated”. The theory for solving linear equations is very well
developed because linear equations are simple enough to be solvable. Nonlinear equations can usually not be solved exactly and are the subject of
much on-going research.
Now, we say that Equation (2.1) is homogeneous if g(t) ≡ 0 for all a <
t < b. If there is a a < t < b such that g(t) 6= 0 then Equation (2.1) is called
non-homogeneous.
Example 2.1
Classify each of the following first order differential equations as linear or
non-linear. If the equation is linear, decide whether it is homogeneous or
non-homogeneous.
y
+ 10
= ty.
(a) dy
dt
2
2
(b) t − 3y + 2ty dy
= 0.
dt
dy
2
(c) t dt = t − 2y.
t−y
(d) dy
= t+y
.
dt
Solution.
1
(a) Notice that the given equation can be written as dy
+ ( 10
− t)y = 0 which
dt
1
is a homogeneous first order linear DE where p(t) = 10 − t and g(t) = 0.
(b) This is non-linear because of the term y 2 .
(c) This is a non-homogeneous first order linear DE since the right-hand side
is not identically zero on any interval. Here, we have p(t) = 2t and g(t) = t.
(d) This is non-linear because of the y in the denominator
First order linear differential equations possess important linearity or superposition properties.
Theorem 2.1
(a) If y1 (t) and y2 (t) are any two solutions of the homogeneous equation
y 0 +p(t)y = 0 then for any constants c1 and c2 the linear combination c1 y1 (t)+
c2 y2 (t) is also a solution of the homogeneous equation.
(b) If y1 (t) is a solution to the homogeneous equation y 0 + p(t)y = 0 and
15
y2 (t) is a solution to the non-homogeneous equation y 0 + p(t)y = g(t) then
Cy1 (t) + y2 (t) is also a solution to the non-homogeneous equation, where C
is an arbitrary constant.
Proof.
(a) Since y1 (t) and y2 (t) are solutions to the homogeneous equation, we have
(c1 y1 + c2 y2 )0 + p(t)(c1 y1 + c2 y2 ) = c1 (y10 + p(t)y1 ) + c2 (y20 + p(t)y2 ) = 0 + 0 = 0.
(b) We have
(Cy1 + y2 )0 + p(t)(Cy1 + y2 ) = C(y10 + p(t)y1 ) + y20 + p(t)y2 = 0 + g(t) = g(t)
Remark 2.1
Part (a) of the previous theorem is not true in general for non-homogeneous equations. For example, consider the equation y 0 = 1. Then y1 (1) = t and y2 (t) = t + 1
are both solutions to the DE. However, y1 (t) + y2 (t) = 2t + 1 is not a solution since
(y1 + y2 )0 = 2 6= 1
Next, we look at the conditions that guarantee the existence of a unique solution
to the IVP
y 0 + p(t)y = g(t), y(t0 ) = y0 .
(2.2)
Theorem 2.2
If p(t) and g(t) are continuous functions in the open interval I = (a, b) and t0 a
point inside I then the IVP (2.2) has a unique solution y(t) defined on I.
Example 2.2
If p(t) is continuous on (a, b) and a < t0 < b then what is the unique solution to
the IVP
y 0 + p(t)y = 0, y(t0 ) = 0?
Solution.
Since the trivial solution y(t) ≡ 0 satisfies the initial value problem, by Theorem
2.2, y(t) ≡ 0 is the unique solution
Remark 2.2
Theorem 2.2 asserts that if the hypotheses are satisfied then a unique solution
exists on an interval containing t0 . However, the solution might actually exist on
a larger interval than what the theorem asserts. To be more precise, consider the
IVP ty 0 + 2y = 0, y(1) = 0. If we apply Theorem 2.2, then a unique solution exists
say on the interval 0 < t < ∞ since this is the interval containing 1 and where
p(t) = 1t is defined and continuous. Actually, the solution is y(t) ≡ 0. But one can
easily see that y(t) ≡ 0 is a solution for all −∞ < t < ∞. So our theorem asserts
the existence of a solution locally rather than globally
162 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP
Practice Problems
Problem 2.1
2
Find p(t) and y0 so that the function y(t) = 3et is the solution to the IVP
y 0 + p(t)y = 0, y(0) = y0 .
Problem 2.2
For each of the initial conditions, determine the largest interval a < t < b on which
Theorem 2.2 guarantees the existence of a unique solution.
y0 +
(a) y(0) = π
t2
1
y = sin t
+1
(b) y(π) = 0.
Problem 2.3
For each of the initial conditions, determine the largest interval a < t < b on which
Theorem 2.2 guarantees the existence of a unique solution
y0 +
(a) y(5) = 2
(b) y(− 32 ) = 1
t
et
y
=
t2 − 4
t−3
(c) y(−6) = 2.
Problem 2.4
(a) For what values of the constant C and the exponent r is y = Ctr the solution
of the IVP
2ty 0 − 6y = 0, y(−2) = 8?
(b) Determine the largest interval of the form a < t < b on which Theorem 2.2
guarantees the existence of a unique solution.
(c) What is the actual interval of existence for the solution found in part (a)?
Problem 2.5
What information does Theorem 2.2 give about the initial value problem ty 0 =
y + t3 cos t, y(1) = 1?y(−1) = 1?
Problem 2.6
Consider the following differential equation
(t − 4)y 0 + 3y =
t2
1
.
+ 5t
Without solving, find the interval over which a unique solution is guaranteed for
each of the following initial conditions:
(a) y(−3) = 4 (b) y(1.5) = −2 (c) y(−6) = 0 (d) y(4.1) = 3.
17
Problem 2.7
2
Without solving the initial value problem, (t − 1)y 0 + (ln t)y = t−3
, y(t0 ) = y0 ,
state whether or not a unique solution is guaranteed to exist for the y(t0 ) = y0
listed below. If a unique solution is guaranteed, find the largest interval for which
the solution satisfies the differential equation and the initial condition.
(a) y(2) = 4 (b) y(0) = 0 (c) y(4) = 2.
Problem 2.8
(a) State precisely the theorem (hypothesis and conclusion) for existence and
uniqueness of a linear first order initial value problem.
3
(b) Consider the equation y 0 + t2 y = et with initial conditions y(t0 ) = y0 . Briefly
discuss if this has a solution, if it is unique and why.
Problem 2.9
Is the differential equation linear or non-linear? If the equation is linear, decide
whether it is homogeneous or non-homogeneous.
(a) y 0 = ty 2
(b) y 0 = t2 y
(c) (cos t)y 0 + et y = sin t
0
(d) yy + t3 = sin t, y > 0.
Problem 2.10
Consider the initial value problem
y 0 + p(t)y = g(t), y(3) = 1.
Suppose that p(t) and/or g(t) have discontinuities at t = −2, t = 0, and t = 5 but
are continuous for all other values of t. What is the largest interval (a, b) on which
Theorem 2.2 is applied.
Problem 2.11
Determine α and y0 so that the graph of the solution to the initial-value problem
y 0 + αy = 0, y(0) = y0
passes through the points (1, 4) and (3, 1).
Problem 2.12
Match the following equations with the correct description. Every equation matches
exactly one description.
(a) y 0 = 3y − 5t.
∂2y
∂2y
(b) ∂y
t = ∂t2 + ∂x2 .
182 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP
(c) y 0 − y 2 = sin t.
(d) y 0 + 3y = 0.
(i) A partial differential equation.
(ii) A homogeneous first order linear differential equation.
(iii) A non-linear first order differential equation.
(iv) A non-homogenous first order linear differential equation.
Problem 2.13
Consider the differential equation y 0 = −t2 y + sin y. What is the order of this
equation? Is it linear or non-linear?
Problem 2.14
Show that y 0 =
t+y
t
is a linear first order non-homogeneous equation.
3 Analytical Solution: The
Method of Integrating Factor
If a unique solution to the initial value problem (2.2) exists then this solution
is found by using the initial condition in the general solution to the differential
equation to determine the value of the parameter. The purpose of this section is
to find analytically the general solution to the first order linear non-homogeneous
equation
y 0 + p(t)y = g(t)
(3.1)
where p(t) and g(t) are continuous on the open interval a < t < b.
The Method of Integrating Factor
We discuss a technique forR solving equation (3.1).
Since p(t) is continuous, it has
R
an antiderivative namely p(t)dt. Let µ(t) = e p(t)dt . Multiply Equation (3.1) by
µ(t) and notice that the left hand side of the resulting equation is the derivative
of a product. Indeed,
d
(µ(t)y) = µ0 (t)y(t) + µ(t)y 0 (t) = µ(t)(y 0 (t) + p(t)y(t)) = µ(t)g(t).
dt
Integrate both sides of the last equation with respect to t to obtain
Z
µ(t)y = µ(t)g(t)dt + C.
Now, divide through by µ(t), we find
Z
1
C
y(t) =
µ(t)g(t)dt +
µ(t)
µ(t)
or
−
y(t) = e
R
p(t)dt
Z
e
R
p(t)dt
19
g(t)dt + Ce−
R
p(t)dt
.
203 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR
One
the form y(t) = Cy1 (t) + y2 (t) where y1 (t) =
R can write the aboveRfunction
R Rinp(t)dt
−
p(t)dt
−
p(t)dt
g(t)dt. Notice that y1 is a solution for the
e
and y2 (t) = e
e
homogeneous equation
y 0 + p(t)y = 0
and y2 is a solution to the non-homogeneous equation as shown in the next example. Thus, the general solution to Equation (3.1) is the sum of a particular
solution of the non-homogeneous equation and the general solution of the homogeneous equation. This solution structure will appear again when discussing second
order linear equations.
Remark 3.1
Notice that multiplying Equation (3.1) by µ(t) was a key factor in the integration
step discussed above. That’s why µ(t) is known as the integrating factor.
Example 3.1
R
R R
Show that yp = e− p(t)dt e p(t)dt g(t)dt satisfies Equation (3.1).
Solution.
We have
yp0
+ p(t)yp = − p(t)e
+p(t)e−
R
−
R
p(t)dt
p(t)dt
Z
R
e
Z
e
R
p(t)dt
p(t)dt
g(t)dt + e−
R
p(t)dt
·e
R
p(t)dt
g(t)
g(t)dt
=g(t)
Example 3.2
Solve the initial-value problem
y0 −
y
= 4t, y(1) = 5.
t
Solution.
By Theorem 2.2, the solution exists and is unique on the interval (0, ∞) since 1
belongs to that interval.
We have p(t) = − 1t so that µ(t) = 1t . Multiplying the given equation by the
integrating factor and using the product rule we notice that
1
y
t
0
= 4.
21
Integrating with respect to t and then solving for y we find that the general solution
is given by
Z
4dt + Ct = 4t2 + Ct.
y(t) = t
The condition y(1) = 5 implies C = 1 and hence the unique solution to the IVP
is y(t) = 4t2 + t, 0 < t < ∞
Example 3.3
Find the general solution to the equation
2
y 0 + y = ln t, t > 0.
t
Solution.
R 2
The integrating factor is µ(t) = e t dt = t2 . Multiplying the given equation by t2
to obtain
(t2 y)0 = t2 ln t.
Integrating with respect to t we find
Z
t2 y = t2 ln tdt + C.
The integral on the right-hand side is evaluated using integration by parts with
3
u = ln t, dv = t2 dt, du = dtt , v = t3 obtaining
t2 y =
t3
t3
ln t − + C.
3
9
Thus,
y=
t
t
C
ln t − + 2
3
9 t
Remark 3.2
Instead of using indefinite integrals
R in the above
R t discussion one can use definite
integrals. For example, replace p(t)dt by t0 p(s)ds for some fixed t0 . Using
definite integral is proven to be useful when p(t) does not have an elementary
function as an antiderivative. For example, when p(t) = sint t or p(t) = sin (t2 ). We
illustrate this idea in the next example.
Example 3.4
Solve y 0 − 1t y = sin t, y(1) = 3. Express your answer in terms of the sine integral,
Rt
Si(t) = 0 sins s ds.
223 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR
Solution.
Since p(t) = − 1t we find µ(t) = 1t . Thus,
0
Z t
0
1
sin t
sin s
y =
=
ds
t
t
s
0
1
y(t) =Si(t) + C
t
y(t) =tSi(t) + Ct.
Since y(1) = 3, we find C = 3 − Si(1). Hence, y(t) = tSi(t) + (3 − Si(1))t
Case when either p(t) or g(t) has a jump discontinuity
Consider the IVP
y 0 + p(t)y = g(t), y(a) = y0 , a ≤ t ≤ b
(3.2)
where p(t) and g(t) are continuous in a ≤ t ≤ b except at t = c where either p(t)
or g(t) has a jump discontinuity at a < c < b. Although y 0 is not continuous at c,
we seek a solution y(t) that is continuous at t = c.
To solve this problem, we first solve the initial value problem on the interval
a ≤ t < c where both p(t) and g(t) are continuous. Let y1 (t) be the unique
solution. Since we are seeking a continuous solution to (3.2), we expect y1 (t) to
have a one-sided limit at c, i.e.,
lim y1 (t) = y1 (c− ).
t→c−
Next, we find the unique solution y2 (t) to the IVP
y 0 + p(t)y = g(t), y(c) = y1 (c− )
where c ≤ t ≤ b. The unique solution to the original IVP is then given by
y1 (t), if a ≤ t < c
y(t) =
y2 (t) if c ≤ t ≤ b.
We illustrate this process in the next example.
Example 3.5
Find the solution to the IVP
1
y 0 + y = g(t), y(1) = 1
t
where
g(t) =
3t, if 1 ≤ t ≤ 2
0 if 2 < t ≤ 3.
The graph of g(t) is given in Figure 3.1(a).
23
Solution.
First, we solve the IVP
1
y 0 + y = 3t, y(1) = 1, 1 ≤ t ≤ 2.
t
The integrating factor is µ(t) = t and the general solution is y1 (t) = t2 +
y(1) = 1, we have C = 0. Hence, y1 (t) = t2 and y1 (2) = 4.
Next, we solve the IVP
C
t.
Since
C
t.
Since
1
y 0 + y = 0, y(2) = 4, 2 < t ≤ 3.
t
The integrating factor is µ(t) = t and the general solution is y2 (t) =
y2 (2) = 4 we find C = 8. Thus,
2
t , if 1 ≤ t ≤ 2
y(t) =
8
if 2 < t ≤ 3.
t
The graph of y(t) is given in Figure 3.1(b). As you can see from the graph, y(t) is
continuous on [1, 3] but not differentiable at t = 2
Figure 3.1
243 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR
Practice Problems
Problem 3.1
Solve the IVP
y 0 = −2ty, y(1) = 1.
Problem 3.2
Solve the IVP
y 0 + et y = 0, y(0) = 2.
Problem 3.3
Consider the first order linear non-homogeneous IVP
y 0 + p(t)y = αp(t), y(t0 ) = y0 .
(a) Show that the IVP can be reduced to a first order linear homogeneous IVP by
the change of variable z(t) = y(t) − α.
(b) Solve this initial value problem for z(t) and use the solution to determine y(t).
Problem 3.4
Apply the results of the previous problem to solve the IVP
y 0 + 2ty = 6t, y(0) = 4.
Problem 3.5
The unique solution to the IVP
ty 0 − αy = 0, y(1) = y0
goes through the points (2, 1) and (4, 4). Find the values of α and y0 .
Problem 3.6
The table below lists values of t and ln [y(t)] where y(t) is the unique solution to
the IVP
y 0 + tn y = 0, y(0) = y0 .
t
ln [y(t)]
1
−0.25
2
−4.00
(a) Determine the values of n and y0 .
(b) What is y(−1)?
3
−20.25
4
−64.00
25
Problem 3.7
Consider the differential equation y 0 + p(t)y = 0. Find p(t) so that y =
general solution.
c
t
is the
Problem 3.8
Consider the differential equation y 0 + p(t)y = 0. Find p(t) so that y = ct3 is the
general solution.
Problem 3.9
Solve the initial-value problem: y 0 − 3t y = 0, y(2) = 8.
Problem 3.10
Solve the differential equation y 0 − 2ty = 0.
Problem 3.11
Find the function f (t) that crosses the point (0, 4) and whose slope satisfies f 0 (t) =
2f (t).
Problem 3.12
Find the general solution to the differential equation y 00 −2y 0 = 0. Hint: Let y 0 = z.
Problem 3.13
Solve the IVP: y 0 + 2ty = t, y(0) = 0.
Problem 3.14
Find the general solution: y 0 + 3y = t + e−2t .
Problem 3.15
Find the general solution: y 0 + 1t y = 3 cos t, t > 0.
Problem 3.16
Find the general solution: y 0 + 2y = cos (3t).
Problem 3.17
Given that the solution to the IVP ty 0 + 4y = αt2 , y(1) = − 31 exists on the interval
−∞ < t < ∞. What is the value of the constant α?
Problem 3.18
Suppose that y(t) = Ce−2t +t+1 is the general solution to the equation y 0 +p(t)y =
g(t). Determine the functions p(t) and g(t).
263 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR
Problem 3.19
Suppose that y(t) = −2e−t + et + sin t is the unique solution to the IVP y 0 + y =
g(t), y(0) = y0 . Determine the constant y0 and the function g(t).
Problem 3.20
Find the value (if any) of the unique solution to the IVP y 0 + (1 + cos t)y =
1 + cos t, y(0) = 3 in the long run, i.e., limt→∞ y(t)?
Problem 3.21
Find the solution to the IVP
y 0 + p(t)y = 2, y(0) = 1
where
p(t) =
0,
1
t
if 0 ≤ t ≤ 1
if 1 < t ≤ 2.
Problem 3.22
Find the solution to the IVP
y 0 + (sin t)y = g(t), y(0) = 3
where
g(t) =
sin t,
if 0 ≤ t ≤ π
− sin t if π < t ≤ 2π.
Problem 3.23
Find the solution to the IVP
y 0 + y = g(t), t > 0, y(0) = 3
where
g(t) =
1, if 0 ≤ t ≤ 1
0
if t > 1.
Problem 3.24
Find the solution to the IVP
y 0 + p(t)y = 0, y(0) = 3
where

 2t − 1, if 0 ≤ t ≤ 1
0,
if 1 < t ≤ 3
p(t) =

if 3 < t ≤ 4.
− 1t
Sketch an accurate graph of the solution, and discuss the long-run behavior of the
solution. Is the solution differentiable on the interval (0, ∞)? Explain your answer.
27
Problem 3.25
Solve the initial-value problem y 0 + y = et y 2 , y(0) = 1 using the substitution
1
u(t) = y(t)
.
283 ANALYTICAL SOLUTION: THE METHOD OF INTEGRATING FACTOR
4 Existence and Uniqueness of
Solutions to First Order
Nonlinear IVP
When a mathematical model is constructed for physical systems, two reasonable
demands are made. First, solutions should exist if the model is to be useful at
all. Second, to work effectively in predicting the future behavior of the physical
system, the model should produce only one solution for a particular set of initial
conditions. Existence and uniqueness theorems help to meet these demands.
In this section we discuss the conditions that guarantee the existence of a unique
solution to the initial value problem
y 0 = f (t, y), y(t0 ) = y0
(4.1)
where f (t, y) needs not be linear in y. Before we proceed to the major result of
this section, we remind the reader of the definition of a partial derivative.
Partial Derivatives
If f (t, y) is a function of two variables t and y then the partial derivative ∂f
∂t of
f (t, y) is the derivative of f (t, y) with respect to t, while pretending y is a constant. The partial derivative ∂f
∂y is the derivative of f (t, y) with respect to y, while
pretending t is constant. The precise definitions are
∂f
∂t (t, y)
= limh→0
f (t+h,y)−f (t,y)
h
and
∂f
∂y (t, y)
= limh→0
Example 4.1
∂f
4 3
5
Find ∂f
∂t and ∂y if f (t, y) = t y + t .
Solution.
We have
∂f
∂t (t, y)
= 4t3 y 3 + 5t4 and
29
∂f
∂y (t, y)
= 3t4 y 2
f (t,y+h)−f (t,y)
.
h
304 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IV
The major existence and uniqueness theorem is stated next.
Theorem 4.1
Suppose that f (t, y) and
∂f
∂y (t, y)
are continuous on the open rectangle
R = {(t, y) : a < t < b, c < y < d}.
Then for any (t0 , y0 ) in R the IVP
y 0 = f (t, y), y(t0 ) = y0
has a unique solution defined on an interval of the form [t0 − h, t0 + h] ⊂ [a, b] for
some positive h.
Remark 4.1
Although existence can be proved with no hypotheses on f beyond continuity,
some assumption such as the continuity of ∂f
∂y is necessary for uniqueness. For
example, the IVP
1
y 0 = y 3 , y(0) = 0
has as solutions the functions y(t) ≡ 0 and
y(t) =
Note that
∂f
∂y (t, y)
if t ≤ 0
if t > 0.
0
3
( 23 t) 2
2
= 13 y − 3 is not continuous at y = 0
Remark 4.2
If the conditions of Theorem 4.1 are not satisfied then this does not mean that the
problem does not have a unique solution.
Example 4.2
Consider the differential equation
1
y3
.
y =
t(y − 2)
0
Does the existence theorem guarantee the existence of a unique solution to the
following IVPs: (a) y(3) = 4 (b) y(0) = 7 (c) y(0) = 2 (d) y(1) = 2?
31
Solution.
The function f (t, y) =
1
y3
t(y−2)
is continuous for t 6= 0 and y 6= 2. The function
∂f
−2 − 2y
(t, y) =
2
∂y
3t(y − 2)2 y 3
is continuous for t 6= 0 and y 6= 0, 2. Thus, f and ∂f
∂y are continuous for t 6= 0
and y 6= 0, 2. Theorem 4.1 guarantees the existence of a unique solution for the
initial value problem in (a) whereas there is no guarantee that there is a unique
solution− or any solution− for the remaining IVPs
Example 4.3
On what rectangle do we expect a unique solution to the IVP
y0 =
y2
, y(0) = 0?
1 − t2
Solution.
We have
f (t, y) =
y2
1 − t2
and
∂f
2y
(t, y) =
.
∂y
1 − t2
These are both continuous functions as long as we avoid the lines t = ±1. Theorem
4.1 tells us that we can expect one and only one solution of
y0 =
y2
, y(0) = 0
1 − t2
in the strip
R = {(t, y) : −1 < t < 1, −∞ < y < ∞}
Example 4.4
Consider the IVP
p
1
y 0 = (−t + t2 + 4y), y(2) = −1.
2
2
(a) Show that y(t) = 1 − t and y(t) = − t4 are two solutions to the above IVP.
(b) Does this contradict Theorem 4.1?
Solution.
(a) You can verify that the
ptwo functions are solutions by substitution.
(b) Since f (t, y) = 12 (−t+ t2 + 4y) and fy (t, y) = √ 21 , these two functions are
t +4y
not continuous at (2, −1). Thus, we can not apply Theorem 4.1 for this problem
324 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IV
Example 4.5
Consider the initial value problem: t2 y 0 − y 2 = 0, y(1) = 1.
(a) Determine the largest open rectangle in the ty−plane, containing the point
(t0 , y0 ) = (1, 1), in which the hypotheses of Theorem 4.1 are satisfied.
(b) A solution of the initial value problem is y(t) = t. This solution exists on
−∞ < t < ∞. Does this fact contradicts Theorem 4.1? Explain your answer.
Solution.
(a) We have f (t, y) =
y2
,
t2
fy (t, y) =
2y
.
t2
So
R = {(t, y) : 0 < t < ∞, − ∞ < y < ∞}.
(b) No. Theorem 4.1 is a local existence theorem and not a global one
Example 4.6
Is it possible to find a function f (t, y) that is continuous and has continuous partial
derivatives in the entire ty−plane such that the functions y1 (t) = cos t and y2 (t) =
1 − sin t are both solutions to the initial value problem y 0 = f (t, y), y π2 = 0?
Solution.
Since f is continuous and has continuous partial derivatives in the entire ty-plane,
the equation y 0 = f (t, y) satisfies the conditions of Theorem 4.1. Notice that
y1 ( π2 ) = y2 ( π2 ) = 0, so the curves y1 (t) = cos t and y2 (t) = 1 − sin t have a common
point ( π2 , 0), so if they were both solutions of our equation, by the uniqueness
theorem they would have to agree on any rectangle containing ( π2 , 0). Since they
do not, they cannot both be solutions of the equation y 0 = f (t, y)
33
Practice Problems
For the given initial value problem in Problems 4.1 - 4.5,
(a) Rewrite the differential equation, if necessary, to obtain the form
y 0 = f (t, y), y(t0 ) = y0 .
Identify the function f (t, y).
∂f
(b) Compute ∂f
∂y . Determine where in the ty−plane both f (t, y) and ∂y are continuous.
(c) Determine the largest open rectangle in the ty−plane that contains the point
(t0 , y0 ) and in which the hypotheses of Theorem 4.1 are satisfied.
Problem 4.1
π
3y 0 + 2t cos y = 1, y( ) = −1.
2
Problem 4.2
π
3ty 0 + 2 cos y = 1, y( ) = −1.
2
Problem 4.3
2t + (1 + y 3 )y 0 = 0, y(1) = 1.
Problem 4.4
(y 2 − 9)y 0 + e−y = t2 , y(2) = 2.
Problem 4.5
cos yy 0 = 2 + tan t, y(0) = 0.
Problem 4.6
Give an example of an initial value problem for which the open rectangle
R = {(t, y) : 0 < t < 4, −1 < y < 2}
represents the largest region in the ty−plane where the hypotheses of Theorem 4.1
are satisfied.
344 EXISTENCE AND UNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IV
Problem 4.7
Can we apply Theorem 4.1 to the following problem ? Explain what (if anything)
we can conclude, and why (or why not):
y
y 0 = √ , y(0) = 2.
t
Problem 4.8
Consider the differential equation y 0 = t−y
t+y . For which of the following initial value
conditions does Theorem 4.1 apply?
(a) y(0) = 0 (b) y(1) = −1 (c) y(−1) = −1.
Problem 4.9
Does the initial value problem y 0 =
Theorem 4.1?
y
t
+ 2, y(0) = 1 satisfy the conditions of
Problem 4.10
Does the initial value problem y 0 = y sin y + t, y(0) = −1 satisfy the conditions of
Theorem 4.1?
5 Separable Differential
Equations
In this section, we discuss an analytical method for solving a type of nonlinear
first order differential equations, the separable differential equations.
A first order differential equation is separable if it can be written with one variable
only on the left and the other variable only on the right:
f (y)y 0 = g(t).
To solve this equation, we proceed as follows. Let F (t) be an antiderivative of f (t)
and G(t) be an antiderivative of g(t). Then by the Chain Rule
d
dF dy
F (y) =
= f (y)y 0 .
dt
dy dt
Thus,
f (y)y 0 − g(t) =
d
d
d
F (y) − G(t) = [F (y) − G(t)] = 0.
dt
dt
dt
It follows that
F (y) − G(t) = C
which is equivalent to
Z
0
Z
f (y)y dt =
g(t)dt + C.
As you can see, the result is generally an implicit equation involving a function of
y and a function of t. It may or may not be possible to solve this to get y explicitly
as a function of t. For an initial value problem, substitute the values of t and y by
t0 and y0 to get the value of C.
Remark 5.1
If F is a differentiable function of y and y is a differentiable function of t and both
F and y are given then the chain rule allows us to find dF
dt given by
dF
dF dy
=
· .
dt
dy dt
35
36
5 SEPARABLE DIFFERENTIAL EQUATIONS
For separable equations, we are given f (y)y 0 = dF
dt and we are asked to find F (y).
This process is referred to as “reversing the chain rule.”
Example 5.1
Solve the initial value problem y 0 = 6ty 2 , y(1) =
1
25 .
Solution.
Since f (t, y) = 6ty 2 and fy (t, y) = 12ty are continuous in the rectangle
R = {(t, y) : −∞ < t < ∞, − ∞ < y < ∞}
1
and R contains the point 1, 25
, by Theorem 4.1, the IVP has a unique solution
on some interval containing t = 1.
Separating the variables and integrating both sides we obtain
Z 0
Z
y
dt = 6tdt
y2
or
−
Z 0
Z
1
dt = 6tdt.
y
Thus,
−
Since y(1) =
explicitly by
1
25 ,
1
= 3t2 + C.
y(t)
we find C = −28. The unique solution to the IVP is then given
y(t) =
1
.
28 − 3t2
The next question is the question of the interval of existence of this solution. Recall
that there are two conditions that define an interval of validity. First, it must be a
continuous interval with no breaks or holes in it. Second it must contain the value
of the independent variable in the initial condition, t = 1 in this case.
There are three possible intervals where y(t) is continuous:
r
−∞ < t < −
28
,
3
r
−
28
<t<
3
r
28
, t>
3
r
28
.
3
Only one of these will contain the value of t from the initial condition and so we
can see that
r
r
28
28
−
<t<
3
3
37
must be the interval of existence for this solution. Figure 5.1 shows the graph of
the solution
Figure 5.1
Example 5.2
Solve the IVP yy 0 = 4 sin (2t), y(0) = 1.
Solution.
This is a separable differential equation. Integrating both sides we find
Z 2 0
Z
y
dt = 4 sin (2t)dt.
2
Thus,
y 2 = −4 cos (2t) + C.
Since y(0) = 1, we find C = 5. Now, solving explicitly for y(t) we find
√
y(t) = ± −4 cos t + 5.
√
Since y(0) = 1, we find y(t) = −4 cos t + 5. The interval of existence of the
solution is the interval −∞ < t < ∞
Example 5.3
Solve the initial value problem
y0 =
p
1 − y 2 , y(0) = 0.
Solution.
Separating the variables and then integrating we find
Z
Z
y0
p
dt = dt
1 − y2
38
5 SEPARABLE DIFFERENTIAL EQUATIONS
or
arcsin y = t + C.
Since y(0) = 0, we find C = 0 and consequently y(t) = sin t where − π2 ≤ t ≤ π2 .
Now, notice that y1 (t) = 1 and y1 (t) = −1 are solutions to the differential equations. Moreover, y( π2 ) = 1 and y(− π2 ) = −1 so that the graph of y = arcsin t is
connected to the solution y1 at the point t = π2 and to the solution y2 at the point
t = − π2 . Thus, the solution to the given IVP is

 −1, −∞ < t < − π2
sin t, − π2 ≤ t ≤ π2
y(t) =

π
1
2 < t < ∞.
The graph of this function is shown in Figure 5.2
Figure 5.2
Linear Versus Nonlinear Differential Equations
Next, we turn our attention to some general questions about differential equations
and explore in more detail some important ways in which nonlinear differential
equations differ from linear ones.
1. Existence and Uniqueness of Solutions
If the functions p(t) and g(t) for a linear differential equation are continuous on
an interval a < t < b containing the initial condition t0 then Theorem 2.2 asserts
the existence of a unique solution throughout the interval a < t < b. This interval
can be identified by glancing at the functions p(t) and g(t).
Turning now to nonlinear differential equations, the function f (t, y) = 6ty 2 in Example 5.1 is continuous everywhere however the intervalqof existence
q of solution
to the initial value problem y 0 = 6ty 2 and y(1) =
1
25
is −
cannot predict this by simply looking at the equation
y0
=
28
3
<t<
28
3
and one
6ty 2 .
2. General Solutions: There may not be a single formula that gives all solutions
39
For a first order linear differential equation the general solution involves a constant from which all particular solutions are obtained by specifying values for this
constant. In contrast, this is not the case for nonlinear differential equations.
If we consider the initial value problem y 0 = 6ty 2 , y(0) = 0 then we see that the
family of solutions y(t) = 3t−1
2 +C does not include the zero solution. Indeed, there
is no choice of C that yields the unique solution y ≡ 0 guaranteed by Theorem
4.1. Such a solution is called a singular solution.
3. Implicit Solutions
Recall again that, for an initial value problem of the form y 0 +p(t)y = g(t), y(t0 ) =
y0 , the unique solution is always given explicitly. In contrast, the situation for nonlinear differential equations is much less satisfactory. For example, the solutions
in Examples 5.1 - 5.3, are defined explicitly. However, the unique solution to the
initial value problem (2y − sin y)y 0 = sin t − t, y(0) = 0 is defined by the implicit
2
equation y 2 + cos y + cos t + t2 = 2 (See Problem 5.11) where y can not be written
explicitly as a function of t.
40
5 SEPARABLE DIFFERENTIAL EQUATIONS
Practice Problems
Problem 5.1
Solve the (separable) differential equation
2 −ln y 2
y 0 = tet
.
Problem 5.2
Solve the (separable) differential equation
y0 =
t2 y − 4y
.
t+2
Problem 5.3
Solve the (separable) differential equation
ty 0 = 2(y − 4).
Problem 5.4
Solve the (separable) differential equation
y 0 = 2y(2 − y).
Problem 5.5
Solve the IVP:
y0 =
Problem 5.6
Solve the IVP:
4 sin (2t)
, y(0) = 1.
y
π
yy 0 = sin t, y( ) = −2.
2
Problem 5.7
Solve the IVP:
y0 +
1
= 0, y(1) = 0.
y+1
Problem 5.8
Solve the IVP:
y 0 − ty 3 = 0, y(0) = 2.
Problem 5.9
Solve the IVP:
π
y 0 = 1 + y 2 , y( ) = −1.
4
41
Problem 5.10
Solve the IVP:
1
y 0 = t − ty 2 , y(0) = .
2
Problem 5.11
Solve the IVP:
(2y − sin y)y 0 = sin t − t, y(0) = 0.
Problem 5.12
1
For what values of the constants α, y0 , and integer n is the function y(t) = (4+t)− 2
a solution of the initial value problem
y 0 + αy n = 0, y(0) = y0 ?
Problem 5.13
State an initial value problem, with initial condition imposed at t0 = 2, having
implicit solution y 3 + t2 + sin y = 4.
Problem 5.14
Consider the initial value problem
y 0 = 2y 2 , y(0) = y0 .
For what value(s) of y0 will the solution have a vertical asymptote at t = 4, where
the t−interval of existence is −∞ < t < 4?
Problem 5.15
Assume that y sin y − 3t + 3 = 0 is an implicit solution of the initial value problem
y 0 = f (y), y(1) = 0. What is f (y)? What is an implicit solution to the initial
value problem y 0 = t2 f (y), y(1) = 0?
Problem 5.16
Find all the solutions to the differential equation y 0 =
2ty
1+t .
Problem 5.17
Solve the initial-value problem y 0 = cos2 y cos2 t, y(0) = π4 .
Problem 5.18
Solve the initial-value problem y 0 = et+y , y(0) = 0 and determine the interval on
which the solution y(t) is defined.
42
5 SEPARABLE DIFFERENTIAL EQUATIONS
Problem 5.19
(a) Solve the initial-value problem
y0 =
t2
ey
−
.
e−y
t2
State the name of the method you are using.
(b) Find the solution which satisfies the condition y(1) = 1.
6 Exact Differential Equations
We shall now present another technique for solving first order, non-linear, ordinary
differential equations. This technique is a generalization of the one we used for
separable equations.
We have seen that the solution procedure of separable equations consists of reversing the chain rule. This same procedure works for exact equations but this time
the chain rule is for functions of two variables.
The Extended Chain Rule
You recall the chain rule for functions of one variable: If u is differentiable at
t and f is differentiable at u(t) then the composite function y = f (u(t)) is also
differentiable at t with derivative given by
dy
dy du
=
·
.
dt
du dt
Example 6.1
√
Find the derivative of the function y = e t .
Solution. √
Let u(t) = t and f (t) = et . Then
du
dt
=
1
√
2 t
and
dy
du
= eu . Hence,
√
1
e t
dy
= eu √ = √
dt
2 t
2 t
The above chain rule can be extended to functions of two variables. Suppose that
u and v are differentiable at t and f is a differentiable function of u and v. Then
the function z(t) = f (u(t), v(t)) is differentiable at t with derivative
∂f du ∂f dv
dz
=
+
.
dt
∂u dt
∂v dt
Example 6.2
Let z = f (u, v) = u2 + 2u − uv + v 2 where u(t) = t2 + 1 and v(t) = t3 − t2 . Find
dz dt t=2 in two different ways.
43
44
6 EXACT DIFFERENTIAL EQUATIONS
Solution.
First notice that u(2) = 5 and v(2) = 4. By using the extended chain rule we have
dz ∂f du ∂f dv
=
+
dt ∂u dt
∂v dt
=(2u + 2 − v)(2t) + (2v − u)(3t2 − 2t).
Thus,
dz = (10 + 2 − 4)(4) + (8 − 5)(8) = 56.
dt t=2
A different way for finding the derivative is to write z as only a function of t
obtaining
z(t) = t6 − 3t5 + 3t4 − t3 + 5t2 + 3.
Finding the derivative of z(t)
z 0 (t) = 6t5 − 15t4 + 12t3 − 3t2 + 10t.
Finally, z 0 (2) = 56
Exact Differential Equations
The basic idea underlying separable equations is to reverse the chain rule for functions of one variable. The basic idea underlying exact equations is to reverse the
extended chain rule. To this end, consider the differential equation
M (t, y) + N (t, y)
dy
= 0.
dt
(6.1)
Let H(t, y) be a function satisfying the two conditions
∂H
∂H
(t, y) = M (t, y) and
(t, y) = N (t, y).
∂t
∂y
(6.2)
Then Equation (6.1) can be written as
∂H
∂H dy
+
= 0.
∂t
∂y dt
By the extended chain rule, Equation (6.3) is the same as
d
H(t, y) = 0.
dt
Therefore, we obtain an implicitly defined solution given by
H(t, y) = C.
(6.3)
45
An equation like (6.1) is called exact if there is a function H(t, y) satisfying the
conditions in (6.2).
Testing a Differentiable Equation for Exactness
The next question is the question of telling whether or not Equation (6.1) is exact.
This is answered by the following theorem.
Theorem 6.1
Suppose that the functions M (t, y) and N (t, y) in (6.1) are continuous and have
∂N
continuous first partial derivatives ∂M
∂y and ∂t in an open rectangle
R = {(t, y) : a < t < b, c < y < d}.
Then (6.1) is exact in R if and only if
∂N
∂M
=
∂y
∂t
for all (t, y) in R.
Remark 6.1
Every separable differential equation is exact. Indeed, since −g(t) + f (y)y 0 = 0
∂g
we have ∂y
= 0 and ∂f
∂t = 0. However, not every exact equation is separable.
For example, the differential equation (2t + y) + (2y + t)y 0 = 0 is exact since
∂M
∂N
∂y = 1 = ∂t = 1. This equation is clearly not separable
Example 6.3
Determine whether or not the equation is exact.
(a) ty 2 + t + t2 yy 0 = 0.
(b) y 2 + 1 + tyy 0 = 0.
(c) cos y + (y 2 + t sin y)y 0 = 0.
(d) cos y + (y 2 − t sin y)y 0 = 0.
Solution.
∂
∂ 2
(a) Since ∂y
(ty 2 + t) = 2ty and ∂t
(t y) = 2ty, the given equation is exact.
∂
∂
2
(b) Since ∂y (y + 1) = 2y and ∂t (ty) = y, the given equation is not exact.
∂
∂
(c) Since ∂y
(cos y) = − sin y and ∂t
(y 2 + t sin y) = sin y, the given equation is not
exact.
∂
∂
(cos y) = − sin y and ∂t
(y 2 − t sin y) = − sin y, the equation is exact
(d) Since ∂y
46
6 EXACT DIFFERENTIAL EQUATIONS
Example 6.4
Consider the initial value problem
t + y + (t + 2y)y 0 = 0, y(0) = 1.
Show that the differential equation is exact and solve the IVP.
Solution.
We have M (t, y) = t + y and N (t, y) = t + 2y. Since
∂M (t, y)
∂N (t, y)
=
=1
∂y
∂t
we have by Theorem 6.1 that the differential equation is exact. Thus,
Z
t2
H(t, y) = (t + y)dt = ty + + c1 (y).
2
Hence,
t + 2y =
It follows that
∂H(t, y)
= t + c01 (y).
∂y
Z
c1 (y) =
(2y)dy = y 2 + C.
Hence,
t2
= C.
2
Since y(0) = 1, we find C = 1. Thus, y satisfies the implicit equation
ty + y 2 +
2ty + 2y 2 + t2 = 2
47
Practice Problems
Problem 6.1
∂f
−ty .
Find ∂f
∂t and ∂y if f (t, y) = y ln y − e
Problem 6.2
∂f
Find ∂f
∂t and ∂y if f (t, y) = ln (ty) +
t2 +1
y−5 .
Problem 6.3
Let f (u, v) = 2u − 3uv where u(t) = 2 cos t and v(t) = 2 sin t. Find
df
dt .
In Problems 6.4 - 6.8, determine whether the given differential equation is exact.
If the equation is exact, find an implicit solution and (where possible) an explicit
solution.
Problem 6.4
yy 0 + 3t2 − 2 = 0, y(−1) = −2.
Problem 6.5
y 0 = (3t2 + 1)(y 2 + 1), y(0) = 1.
Problem 6.6
(6t + y 3 )y 0 + 3t2 y = 0, y(1) = 2.
Problem 6.7
(et+y + 2y)y 0 + (et+y + 3t2 ) = 0, y(0) = 0.
Problem 6.8
(sin (t + y) + y cos (t + y) + t + y)y 0 + (y cos (t + y) + y + t) = 0, y(1) = −1.
Problem 6.9
For what values of the constants m, n, and α (if any) is the following differential
equation exact?
tm y 2 y 0 + αt3 y n = 0.
48
6 EXACT DIFFERENTIAL EQUATIONS
Problem 6.10
Assume that N (t, y)y 0 +t2 +y 2 sin t = 0 is an exact differential equation. Determine
the general form of N (t, y).
Problem 6.11
Assume that t3 y + et + y 2 = 5 is an implicit solution to the differential equation
N (t, y)y 0 + M (t, y) = 0, y(0) = y0 .
Determine possible functions M (t, y), N (t, y), and the possible value(s) for y0 .
Problem 6.12
√
Assume that y = −t − 4 − t2 is an explicit solution of the following initial value
problem
(y + at)y 0 + (ay + bt) = 0, y(0) = y0 .
Determine values for the constants a, b and y0 .
Problem 6.13
Let k be a positive constant. Use the exactness criterion to determine whether or
not the population equation dP
dt = kP is exact. Do NOT try to solve the equation
or carry out any further calculation.
Problem 6.14
Consider the differential equation (2t + 3) + (2y − 2)y 0 = 0. Determine whether
this equation is exact or not. If it is, solve it.
Problem 6.15
Consider the differential equation (ye2ty + t) + bte2ty y 0 = 0. Determine for which
value of b this equation is exact, and then solve it with this value of b.
Problem 6.16
Consider the differential equation y + (2t − yey )y 0 = 0. Check that this equation is
not exact. Now multiply the equation by y. Check that the new equation is exact,
and solve it.
Problem 6.17
(a) Consider the differential equation
y 0 + p(t)y = g(t)
with p(t) 6= 0. Show
that this equation is not exact.
R
p(t)dt
(b) Let µ(t) = e
. Show that the equation
µ(t)(y 0 + p(t)y) = µ(t)g(t)
is exact and solve it.
49
Problem 6.18
Use the method of the previous problem to solve the linear, first-order equation
y 0 − yt = 1, with initial condition y(1) = 7. First, check that this equation is not
exact. Next, find µ(t). Multiply the equation by µ(t) and check that the new
equation is exact. Solve it, using the method of exact equations.
Problem 6.19
Put the following differential equation in the “Exact Differential Equation” form
and find the general solution
y 3 − 2ty
y0 = 2
.
t − 3ty 2
Problem 6.20
The following differential equations are exact. Solve them by that method.
(a) (4t3 y + 4t + 4)y 0 = 8 − 4y − 6t2 y 2 , y(−1) = 1.
(b) (6 − 4y + 16t) + (10y − 4t + 2)y 0 = 0, y(1) = 2.
50
6 EXACT DIFFERENTIAL EQUATIONS
7 Substitution Techniques:
Bernoulli and Riccati Equations
A well-known non-linear equation that reduces to a linear one with an appropriate
substitution is the Bernoulli equation given by
y 0 + p(t)y = g(t)y n
(7.1)
where n is a real number different from 0 and 1. Notice that for n = 0 or n = 1
the equation becomes linear.
To solve (7.1), first notice that y(t) ≡ 0 is the trivial solution. If y(t) 6≡ 0 then
(7.1) can be written as
y0
+ p(t)y 1−n = g(t).
yn
(7.2)
Let z = y 1−n . Then by the chain rule of differentiation we have z 0 = (1 − n)y −n y 0
0
1
and therefore yyn = 1−n
z 0 . Thus, (7.1) reduces to
1
z 0 + p(t)z = g(t)
1−n
which is a first order linear differential equation that can be solved by the technique
1
of integrating factor. Once z is found then the desired solution is y(t) = z 1−n .
Example 7.1
Solve the Bernoulli equation
1
y 0 − y = ty 2 , t > 0.
t
Solution.
Divide by y 2 and then let z = y −1 to obtain
1
z 0 + z = −t.
t
51
527 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS
The integrating factor is µ(t) = t and the general solution is
Z
1
t2
z(t) =
−t2 dt + Ct−1 = − + Ct−1 .
t
3
Thus, the solutions to the differential equation are y =
0
1
z
=
1
2
− t3 +Ct−1
and y(t) =
Example 7.2
Solve the Bernoulli equation y 0 + 3y = e3t y 2 .
Solution.
Dividing by y 2 to obtain
y 2 y 0 + 3y −1 = e3t .
Let z = y −1 . Then,
z 0 − 3z = −e3t .
Solving this equation using the integrating factor method with µ(t) = e−3t we find
Z
3t
z(t) = e
e−3t (−e3t )dt + Ce3t = −te3t + Ce3t = e3t (C − t).
The solutions are y(t) = e−3t (C − t)−1 and y(t) = 0
Example 7.3
Solve: y 0 + 2t y = −t2 y 2 cos t.
Solution.
Dividing through by y 2 to obtain y −2 y 0 + 2t y −1 = −t2 cos t. Letting z = y −1 to
obtain
2
z 0 − z = t2 cos t.
t
Solving this equation by the method of integrating factor with µ(t) =
Z
z(t) = t2 cos tdt + Ct2 = t2 sin t + Ct2 .
The solutions are y(t) = (t2 sin t + Ct2 )−1 and y(t) = 0
1
t2
we find
53
Riccati Equation
A first order differential equation is called a Riccati equation if it can be written
in the form:
y 0 = a(t)y 2 + b(t)y + c(t)
(7.3)
where a, b and c are functions of t. Clearly, this is a first order nonlinear and nonseparable differential equation.
The solution of a Riccati equation requires knowledge of a particular solution to
(7.3). To solve (7.3), we first find a particular solution y1 to (7.3). Then we use
the substitution z1 = y − y1 . Thus, y = z1 + y1 . Now, (7.3) reduces to
−
−
z0
1
y1
1
+ y10 =a(t)( 2 + 2 + y12 ) + b(t)( + y1 ) + c(t)
z2
z
z
z
z0
b(t)
a(t) 2a(t)y1
+ a(t)y12 + b(t)y1 + c(t) = 2 +
+ a(t)y12 +
+ b(t)y1 + c(t)
2
z
z
z
z
z 0 = − a(t) − [2a(t)y1 + b(t)]z.
Thus, a Riccati equation can be reduced to a linear equation
z 0 + [2a(t)y1 + b(t)]z = −a(t)
(7.4)
that can be solved by the method of integrating factor. Once z(t) is found then
1
the solution to the original equation is y(t) = z(t)
+ y1 (t). As the next example
illustrates, in many cases a solution of a Riccati equation cannot be expressed in
terms of elementary functions.
Example 7.4
Solve : y 0 = 2 − 2ty + y 2 given that y1 (t) = 2t.
Solution.
We have a(t) = 1, b(t) = −2t, and c(t) = 2. Substituting in (7.4) to obtain
z 0 + 2tz = −1.
2
The integrating factor is µ(t) = et so that
2
et z
0
2
= −et .
547 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS
Now, the integral
Thus, we write
Rt
2
t0
es ds cannot be expressed in terms of elementary functions.
2
et z(t) = −
Z
t
2
2
es ds + et0 z(t0 ).
t0
Solving for z, we find
z(t) = e
−t2
Z
t20
t
(e z(t0 ) −
2
es ds).
t0
Finally,
y(t) =
1
2
e−t2 (et0 z(t
Example 7.5
Solve the IVP: y 0 = 2 + 2y + y 2 ,
variables.
0)
−
Rt
t0
es2 ds)
+ 2t
y(0) = 0 using the method of separation of
Solution.
Notice first that 2 + 2y + y 2 = 1 + (1 + y)2 . Separating the variables we find
y0
= 1.
1 + (1 + y)2
Integrating both sides with respect to t to obtain
arctan (1 + y) = t + C.
But y(0) = 0 so that C = π4 . Thus,
arctan (1 + y) = t +
π
.
4
For the inverse arc tangent we must have − π2 < t + π4 <
y(t) = tan (t +
π
2
π
or − 3π
4 < t < 4 . Hence,
π
3π
π
) − 1, −
<t<
4
4
4
Example 7.6
Solve the differential equation y 0 = 5 − t2 + 2ty − y 2 given that y1 (t) = t − 2 is a
particular solution.
55
Solution.
Let z1 = y − t + 2. Then the given equation reduces to
z 0 + 4z = 1.
Solving this equation by the method of integrating factor with µ(t) = e4t to obtain
Z
1
z(t) = e−4t e4t dt + Ce−4t = + Ce−4t .
4
Thus,
1
y(t) = ( + Ce−4t )−1 + t − 2
4
567 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS
Practice Problems
Problem 7.1
Solve the Bernoulli equation
y0 =
t2 + 3y 2
, t > 0.
2ty
Problem 7.2
2
Find the general solution of y 0 + ty = te−t y −3 .
Problem 7.3
Solve the IVP ty 0 + y = t2 y 2 , y(0.5) = 0.5.
Problem 7.4
Solve the IVP y 0 − 1t y = −y 2 , y(1) = 1.
Problem 7.5
Solve the IVP y 0 = y(1 − y), y(0) = 21 .
Problem 7.6
Solve y 0 + y = ty 4 .
Problem 7.7
Solve the differential equation y 0 = 1 + t2 − y 2 given that y1 (t) = t is a particular
solution.
Problem 7.8
Perform a change of variable that changes the Bernoulli equation y 0 + y + y 2 = 0
into a linear equation in the new variable. Do NOT try to solve the equation or
proceed further than with any calculations.
Problem 7.9
Consider the equation
y 0 = y − σy 3 , > 0, σ > 0.
(a) Use the Bernoulli transformation to change this nonlinear equation into a linear
equation.
(b) Solve the resulting linear equation in part (a) and use the solution to find the
solution of the given differential equation above.
57
Problem 7.10
Consider the differential equation
y0 = f
y t
.
(a) Show that the substitution z = yt leads to a separable differential equation in
z.
(b) Use the above method to solve the initial-value problem
y0 =
t+y
, y(1) = 0.
t−y
Problem 7.11
Solve: y 0 + y3 = et y 4 .
Problem 7.12
Solve: ty 0 + y = ty 3 .
Problem 7.13
Solve: ty 0 + y = t2 y 2 ln t.
Problem 7.14
Verify that y1 (t) = 2 is a particular solution to the Riccati equation
y 0 = −2 − y + y 2 ,
and then find the general solution.
Problem 7.15
Verify that y1 (t) =
2
t
is a particular solution to the Riccati equation
y0 = −
and then find the general solution.
4
1
− y + y2,
t2
t
587 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS
8 Graphical Solution: Direction
Field of y 0 = f (t, y)
Since explicit solutions of differential equations are often unobtainable, we explore
methods of finding properties of solutions from the differential equation itself; the
principal tool is the geometry of direction field.
A direction field (also known as slope field) consists of an array of short line
segments in the ty-plane having the property that the line plotted at a point
(t, y) has slope f (t, y). Direction fields are basically used to visualize the family of
solutions of a given differential equation without the need of solving the equation.
Direction fields give qualitative information about solutions of ODEs.
In this section we use direction fields for solving initial value problems of the form
dy
= f (t, y), y(t0 ) = y0 .
dt
In the special case where f (t, y) = f (y), i.e. the independent variable t does not
appear on the right side, the first order DE dy
dt = f (y) is called autonomous.
Producing slope fields by hand is a daunting task. For that reason, slope fields are
usually created by means of an electronic generator. One such a generator can be
found at http://slopefield.nathangrigg.net/
Example 8.1
Find the direction field of the differential equation
dy
= 2t.
dt
What is the form of the general solution? Graph the particular solution going
through (0, −1).
Solution.
Figure 8.1 shows the slope field and the graph of the particular solution to the given
59
60
8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y 0 = F (T, Y )
DE passing through the point (0, −1). The solution curves look like parabolas.
Thus, the general solution is given by the equation y = t2 + C
Figure 8.1
Example 8.2
Using direction field, guess the form of the solution curves of the differential equation
dy
t
=− .
dt
y
Solution.
The direction field is given in Figure 8.2. The solution curves look like circles
centered at the origin. Thus, the general solution is given implicitly by the equation
t2 + y 2 = C where C is a positive constant
Figure 8.2
61
Remark 8.1
We point out here that even though one can draw solution curves, some do not
1+yety
have simple formulas. For instance, the equation dy
dt = − 1+tety does not have
explicit solutions.
Equilibrium Solutions and Stability for Autonomous Equations
A physical system is often said to be in equilibrium if it doesn’t change in time. We
adopt this idea and say that a solution to a differential equation is an equilibrium
solution if it is a constant function. Thus, in a direction field of an autonomous
equation, equilibrium solutions are solution curves represented by horizontal lines.
It follows that the equations of such solutions have the form y(t) ≡ c where c is a
constant. The following result tells us where to look for equilibrium solutions.
Theorem 8.1
The function y(t) ≡ c, where c is a constant, is an equilibrium solution to y 0 = f (y)
if and only if c is a root of f (y) = 0.
Example 8.3
Find the equilibrium solutions to the DE
dy
= 2y(1 − y).
dt
Solution.
The roots of f (y) = 2y(1 − y) = 0 are y = 0 and y = 1. According to the previous
theorem, the equilibrium solutions are y(t) ≡ 0 and y(t) ≡ 1. The direction field
of the DE is shown in Figure 8.3
Figure 8.3
62
8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y 0 = F (T, Y )
Remark 8.2
Equilibrium solutions can be defined for nonautonomous differential equations. For
example, the function y(t) ≡ 1 is an equilibrium solution to the DE y 0 = (1 − y)t2 .
The direction field of a given differential equation indicates that as t increases
without bound, every solution either moves towards or moves away from an equilibrium solution.
If all nearby solutions move towards a certain equilibrium solution, then that equilibrium solution is called asymptotically stable, stable, or attracting. The
solution y = 1 in Figure 8.3 is attracting. An equilibrium solution is called unstable or repelling when all nearby solutions move away from it. The solution
y = 0 in Figure 8.3 is repelling.
If solutions on one side of an equilibrium solution move towards the equilibrium
solution and on the other side of the equilibrium solution move away from it then
we call the equilibrium solution semi-stable.
An equilibrium solution does not necessarily have to be either attracting or repelling. The next example illustrates this situation.
Example 8.4
Sketch the field direction of the differential equation y 0 = 4y(1 − y)2 . Show that
y = 1 is neither stable or unstable.
Solution.
The direction field is shown in Figure 8.4. Note that the equilibrium solution
y(t) ≡ 1 is neither stable or unstable. Nearby solutions that start below it are
attracted upward towards it but nearby solutions that start above it are repelled
upward and away from it
Figure 8.4
63
A very suitable qualitative representation of a differential equation for the study
of stability is the so-called phase line. A phase line consists of solid dots and
arrows. The solid dots represent the equilibrium points and the arrows indicate
the directions that solutions move as t increases. Figure 8.5(a) shows an example
of a phase line. We see that the equilibrium b is stable, whereas the equilibria a
and c are unstable.
Example 8.5
Consider the autonomous differential equation dy
dt = f (y) where the graph of f (y)
is given in Figure 8.5(b).
(a) Sketch the phase line.
(b) Sketch the direction field of this differential equation.
(c) Sketch the graph of the solution to the IVP y 0 = f (y), y(0) = 23 . Find
limt→∞ y(t).
(d) Sketch the graph of the solution to the IVP y 0 = f (y), y(0) = 12 . Find
limt→∞ y(t).
(e) Sketch the graph of the solution to the IVP y 0 = f (y), y(0) = − 23 . Find
limt→∞ y(t).
Figure 8.5
Solution.
(a) Note that for y < −1, f (y) < 0 so that the solution y(t) is decreasing. For
−1 < y(t) < 0, we have f (y) > 0 so that y(t) is increasing. For 0 < y(t) < 1, we
have that f (y) < 0 so that y(t) is decreasing. Finally, for y(t) > 1 we have that
f (y) > 0 so that y(t) is increasing. Hence, the phase line is given by Figure 8.5(c).
(b) The direction field is given in Figure 8.6.
(c) We notice from Figure 8.6 that limt→∞ y(t) = ∞.
(d) We notice from Figure 8.6 that limt→∞ y(t) = 0.
(e) We notice from Figure 8.6 that limt→∞ y(t) = −∞
64
8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y 0 = F (T, Y )
Figure 8.6
65
Practice Problems
Problem 8.1
Sketch the direction field for the differential equation in the window −3 ≤ t ≤
3, −3 ≤ y ≤ 3.
(a) y 0 = y (b) y 0 = t − y.
Problem 8.2
Match each direction field with the equation that the slope field could represent. Each direction field is drawn in the portion of the ty-plane defined by
−6 ≤ t ≤ 6, −4 ≤ y ≤ 4.
(a) y 0 = −t (b) y 0 = sin t (c) y 0 = 1 − y (d) y 0 = y(2 − y).
Problem 8.3
State whether or not the equation is autonomous.
(a) y 0 = −t (b) y 0 = sin t (c) y 0 = 1 − y (d) y 0 = y(2 − y).
Problem 8.4
Find all the equilibrium solutions of each of the autonomous differential equations
below
(a) y 0 = (y − 1)(y − 2).
(b) y 0 = (y − 1)(y − 2)2 .
(c) y 0 = (y − 1)(y − 2)(y − 3).
Problem 8.5
Find an autonomous differential equation with an equilibrium solution at y = 1
and satisfying y 0 < 0 for −∞ < y < 1 and 1 < y < ∞.
66
8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y 0 = F (T, Y )
Problem 8.6
Find an autonomous differential equation with no equilibrium solutions and satisfying y 0 > 0.
Problem 8.7
Find an autonomous differential equation with equilibrium solutions y = n2 , where
n is an integer.
Problem 8.8
Find an autonomous differential equation with equilibrium solutions y = 0 and
y = 2 and satisfying the properties y 0 > 0 for 0 < y < 2; y 0 < 0 for y < 0 or y > 2.
Problem 8.9
Classify whether the equilibrium solutions are stable, unstable, or neither.
(a) y 0 = 1 − y 2 .
(b) y 0 = (y + 1)2 .
Problem 8.10
Consider the direction field below. Classify the equilibrium points, as asymptotically stable, semi-stable, or unstable.
Problem 8.11
Sketch the direction field of the equation y 0 = y 3 . Sketch the solution satisfying
the condition y(1) = −1. What is the domain of this solution?
Problem 8.12
Find the equilibrium solutions and determine their stability
y 0 = y 2 (y 2 − 1).
67
Problem 8.13
Find the equilibrium solutions of the equation
y 0 = y 2 − 4y
then decide whether they are stable or unstable. What is the long-run behavior if
y(0) = 5?y(0) = 4?y(0) = 3?
Problem 8.14
What is limt→∞ y(t) for the initial-value problem
y 0 = sin (y(t)), y(0) =
π
?
2
68
8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y 0 = F (T, Y )
9 Numerical Solutions to
ODEs: Euler’s Method and its
Variants
Whenever a mathematical problem is encountered in science or engineering, which
cannot readily or rapidly be solved by a traditional mathematical method, then
a numerical method is usually sought and carried out. In this section, we study
Euler’s method and its variants for approximating the solution to the initial value
problem
y 0 = f (t, y), y(t0 ) = y0 , a < t < b.
(9.1)
Denote the (unique) solution by y(t). Then y(t) satisfies y 0 (t) = f (t, y(t)). A numerical method for solving the initial value problem provides an algorithm for
calculating approximate values y0 , y1 , y2 , · · · , yn , · · · of the solution y(t) at a set of
points t0 < t1 < t2 < · · · < tn < · · · . Usually this is a step-by-step process.
Euler’s Method
Euler’s method is a very simple numerical method for solving the first-order initial
value problem (9.1). Suppose we want to find approximate values y0 , y1 , y2 , · · · , yn , · · ·
of the solution y(t) at equally spaced points t0 < t1 < t2 < · · · < tn < · · · with
step size h. We begin by integrating Equation (9.1) between tn and tn+1 obtaining
Z
tn+1
0
Z
tn+1
y (s)ds =
tn
f (s, y(s))ds.
tn
By the fundamental theorem of calculus we can write
Z
tn+1
y 0 (s)ds = y(tn+1 ) − y(tn ).
tn
69
709 NUMERICAL SOLUTIONS TO ODES: EULER’S METHOD AND ITS VARIANTS
Thus,
Z
tn+1
y(tn+1 ) = y(tn ) +
f (s, y(s))ds.
(9.2)
tn
Computationally, this last equation can not be used since we do not know f (s, y(s))
for tn ≤ s ≤ tn+1 . If we estimate the integral by a left Riemann sum we find
Z tn+1
f (s, y(s))ds ≈ hf (tn , y(tn )).
tn
Hence,
y(tn+1 ) ≈ y(tn ) + hf (tn , y(tn )).
or
yn+1 = yn + hf (tn , yn )
(9.3)
where yn ≈ y(tn ). Equation (9.3) is known as Euler’s method. Therefore, we can
view Euler’s method as a left Riemann sum approximation of the integral equation
(9.2).
Example 9.1
Suppose that y(0) = 1 and
Solution.
The step size is h =
0.5−0
5
dy
dt
= y. Estimate y(0.5) in 5 steps using Euler’s method.
= 0.1. The following chart lists the steps needed:
k
0
1
2
3
4
5
tk
0
0.1
0.2
0.3
0.4
0.5
yk
1
1.1
1.21
1.331
1.4641
1.61051
f (tk , yk )h
0.1
0.11
0.121
0.1331
0.14641
Thus, y(0.5) ≈ 1.61051. Note that the exact value is y(0.5) = e0.5 = 1.6487213
The Improved Euler’s Method: Heun’s Method
Euler’s method is derived from estimating the integral equation by a left Riemann
sum. A better estimate to the integral is obtained by using the trapezoid rule
obtaining
Z tn+1
h
f (s, y(s))ds ≈ [f (tn , y(tn )) + f (tn+1 , y(tn+1 ))].
2
tn
71
This leads to the estimate
yn+1 = yn +
h
[f (tn , yn ) + f (tn+1 , yn+1 )].
2
Note that yn+1 is defined implicitly so solving for yn+1 requires solving a nonlinear
equation say by the Newton’s method for root-finding. Despite the inconvenience
of working with an implicit method, the trapezoid method has two advantages
over the ordinary Euler method: First, it is more accurate and second it is stable
(this means that it is not sensitive to small changes in the initial values).
Heun’s method is based on the evaluation of an estimate of f (tn+1 , yn+1 ) by replacing the value of yn+1 with an estimate derived from the original Euler method,
with the resulting iteration scheme:
ỹn+1 = yn + hf (tn , yn ), yn+1 = yn +
h
[f (tn , yn ) + f (tn+1 , ỹn+1 )]
2
or
h
[f (tn , yn ) + f (tn+1 , yn + hf (tn , yn ))].
2
This equation is known as Heun’s method or the improved Euler’s method.
yn+1 = yn +
Example 9.2
Suppose that y(0) = 1 and
Euler’s method.
Solution.
The step size is h =
0.5−0
5
dy
dt
= y. Estimate y(0.5) in 5 steps using the improved
= 0.1 The following chart lists the steps needed:
y0 =1
y1 =y0 +
y2 =y1 +
y3 =y2 +
y4 =y3 +
y5 =y4 +
h
[f (t0 , y0 ) + f (t1 , y0 + hf (t0 , y0 ))] = 1.105
2
h
[f (t1 , y1 ) + f (t2 , y1 + hf (t1 , y1 ))] = 1.221025
2
h
[f (t2 , y2 ) + f (t3 , y2 + hf (t2 , y2 ))] = 1.349232625
2
h
[f (t3 , y3 ) + f (t4 , y3 + hf (t3 , y3 ))] = 1.490902050625
2
h
[f (t4 , y4 ) + f (t5 , y4 + hf (t4 , y4 ))] = 1.647446765940625
2
Thus, y(0.5) ≈ 1.647446765940625. Note that the exact value is y(0.5) = e0.5 =
1.6487213
729 NUMERICAL SOLUTIONS TO ODES: EULER’S METHOD AND ITS VARIANTS
The Modified Euler’s Method
Another numerical integration scheme is the midpoint rule. In this case, we have
Z tn+1
h
h
f (s, y(s))ds ≈ hf tn + , y tn +
.
2
2
tn
This implies that
h
h
y(tn+1 ) ≈ y(tn ) + hf tn + , y tn +
.
2
2
Using Euler’s method we can write
h
h
≈ y(tn ) + f (tn , y(tn )).
y tn +
2
2
Hence,
yn+1 = yn + hf
h
h
tn + , yn + f (tn , yn ) .
2
2
This last equation is known as the modified Euler’s method.
Example 9.3
Suppose that y(0) = 1 and
Euler’s method.
Solution.
The step size is h =
0.5−0
5
dy
dt
= y. Estimate y(0.5) in 5 steps using the modified
= 0.1. The following chart lists the steps needed:
y0 =1
y1 =y0 + hf (t0 +
y2 =y1 + hf (t1 +
y3 =y2 + hf (t2 +
y4 =y3 + hf (t3 +
y5 =y4 + hf (t4 +
h
, y0 +
2
h
, y1 +
2
h
, y2 +
2
h
, y3 +
2
h
, y4 +
2
h
f (t0 , y0 )) = 1.105
2
h
f (t1 , y1 )) = 1.221025
2
h
f (t2 , y2 )) = 1.349232625
2
h
f (t3 , y3 )) = 1.490902050625
2
h
f (t4 , y4 )) = 1.647446765940625
2
Thus, y(0.5) ≈ 1.647446765940625. Note that the exact value is y(0.5) = e0.5 =
1.6487213
73
Practice Problems
In Problems 9.1 - 9.3, answer the following questions:
(a) Solve the initial value problem analytically, using an appropriate solution technique.
(b) For the given initial value problem express yn+1 in terms of yn using Heun’s
method.
(c) For the given initial value problem express yn+1 in terms of yn using the Modified Euler’s method.
(d) Use a step size h = 0.1. Compute the first three approximations y1 , y2 , y3 using
the method in part (b).
(e) Use a step size h = 0.1. Compute the first three approximations y1 , y2 , y3 using
the method in part (c).
(f) For comparison, calculate and list the exact solution values y(t1 ), y(t2 ), y(t3 ).
Problem 9.1
y 0 = 2t − 1, y(1) = 0.
Problem 9.2
y 0 = −y, y(0) = 1.
Problem 9.3
y 2 y 0 + t = 0, y(0) = 1.
Problem 9.4
Consider the initial value problem
y 0 = 1 + y 2 , y(0) = −1.
(a) Find the exact solution of the given initial value problem.
(b) Use step size h = 0.05. Compute 4 steps of Euler’s method, Heun’s method,
and modified Euler’s method. Compare the numerical values obtained at t = 1 by
calculating the error |y(1) − y4 |.
Problem 9.5
Consider the initial value problem
y 0 + 2y = 4, y(0) = 3.
749 NUMERICAL SOLUTIONS TO ODES: EULER’S METHOD AND ITS VARIANTS
(a) Find the exact solution of the given initial value problem.
(b) Use step size h = 0.05. Compute 4 steps of Euler’s method, Heun’s method,
and modified Euler’s method. Compare the numerical values obtained at t = 1 by
calculating the error |y(1) − y4 |.
In Problems 9.6 - 9.8, the given iteration is the result of applying N steps of Euler’s
method, Heun’s method, or the modified Euler’s method to an initial value problem
of the form
y 0 = f (t, y), y(t0 ) = y0 , t0 ≤ t ≤ t0 + T.
Identify the numerical method and determine t0 , N, T, and f (t, y).
Problem 9.6
yn+1 = yn + h(yn + t2n yn3 ), y0 = 1
tn = 2 + nh, h = 0.02, n = 0, 1, 2, · · · , 49.
Problem 9.7
yn+1
h
h
2
2
= yn + h tn +
sin yn + tn sin yn , y0 = 1
2
2
tn = nh, h = 0.01, n = 0, 1, 2, · · · , 199.
Problem 9.8
h
[tn yn2 + 2 + (tn + h)(yn + h(tn yn2 + 1))2 ], y0 = 2
2
tn = 1 + nh, h = 0.05, n = 0, 1, 2, · · · , 99.
yn+1 = yn +
10 Second Order Linear
Differential Equations:
Existence and Uniqueness
Results
To this point we have considered only first order differential equations. However,
many of the most interesting differential equations involve second derivatives. Indeed, since acceleration is the second derivative of position, Newton’s second law
of motion, F = ma, is a second order differential equation. In this and the coming
sections we turn our attention to linear second-order differential equations.
By a second-order linear differential equation we mean an equation of the
form
y 00 + p(t)y 0 + q(t)y = g(t).
If g(t) ≡ 0 we say that the equation is homogeneous. Otherwise the equation
is non-homogeneous. Initial-value problems for second-order linear differential
equations require two initial-conditions. In this section we will consider the question of existence and uniqueness of solutions to the initial value problem
y 00 + p(t)y 0 + q(t)y = g(t), y(t0 ) = y0 , y 0 (t0 ) = y00 .
(10.1)
Existence and uniqueness results similar to first-order equations exist for secondorder equations as well. The following theorem tells us the conditions for the
existence and uniqueness of solution to (10.1). Note how this theorem is analogous
to the corresponding theorem for first order linear ODE’s with initial conditions.
Theorem 10.1
If p(t), q(t), and g(t) are continuous functions over an interval a < t < b containing
t0 then the initial value problem (10.1) has a unique solution in the interval a <
t < b.
75
7610 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQU
Example 10.1
Find the largest interval where
(t2 − 1)y 00 + 3ty 0 + (cos t)y = et , y(0) = 4, y 0 (0) = 5
is guaranteed to have a unique solution.
Solution.
We first put it into the standard form
y 00 +
cos t
et
3t 0
y
+
y
=
, y(0) = 4, y 0 (0) = 5.
t2 − 1
t2 − 1
t2 − 1
We see that p, q, and g are all continuous except at t = −1 and t = 1. Theorem
10.1 tells us that there is a unique solution on −1 < t < 1 since this interval
contains 0
Remark 10.1
The above theorem does not give the largest t−interval of existence as shown in
the next example. Thus, Theorem 10.1 is a local existence theorem.
Remark 10.2
If the conditions of the above theorem are not guaranteed, this does not mean the
the problem does not have a solution.
Example 10.2
Consider the initial value problem t2 y 00 − ty 0 + y = 0, y(1) = 1, y 0 (1) = 1.
(a) What is the largest interval on which Theorem 10.1 guarantees the existence
of a unique solution?
(b) Show by direct substitution that the function y(t) = t is the unique solution
to this initial value problem. What is the interval on which this solution actually
exists?
(c) Does this example contradict the assertion of Theorem 10.1? Explain.
Solution.
(a) Writing the equation in standard form to obtain
1
1
y 00 − y 0 + 2 y = 0.
t
t
From this, we see that the functions p(t) = − 1t and q(t) = t12 are continuous for all
t 6= 0. Since t0 = 1, the largest t−interval according to Theorem 10.1 is 0 < t < ∞.
(b) If y(t) = t then y 0 (t) = 1 and y 00 (t) = 0 so that y 00 − 1t y 0 + t12 y = 0 − 1t + 1t =
0, y(1) = y 0 (1) = 1. So y(t) = t is a solution so that by Theoren 10.1 it is the only
solution. The t−interval for this solution is −∞ < t < ∞.
(c) No, because the theorem is a local existence theorem and not a global one
77
Example 10.3
Consider the graphs shown. Each graph displays a portion of the solution of one
of the four initial value problems given. Match each graph with the appropriate
initial value problem.
(a) y 00 + y = 2 − sin t, y(0) = 1, y 0 (0) = −1;
(b) y 00 + y = −2t, y(0) = 1, y 0 (0) = −1;
(c) y 00 − y = t2 , y(0) = y 0 (0) = 1;
(d) y 00 − y = −2 cos t, y(0) = y 0 (0) = 1.
Solution.
(a) B since y 0 (0) = −1 < 0 and y 00 (0) = 2 − sin 0 − y(0) = 1 > 0.
(b) D since y 0 (0) = −1 < 0 and y 00 (0) = −2(0) − y(0) = −1 < 0.
(c) A since y 0 (0) = 1 > 0 and y 00 (0) = 02 + y(0) = 1 > 0.
(d) C since y 0 (0) = 1 > 0 and y 00 (0) = −2 cos 0 + y(0) = −1 < 0
Example 10.4
Give an example of an initial value problem of the form y 00 + p(t)y 0 + q(t)y =
0, y(t0 ) = y0 , y 0 (t0 ) = y00 for which the given t−interval −1 < t < 5 is the largest
on which Theorem 10.1 guarantees a unique solution.
Solution.
One such an answer is
7810 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQU
y 00 +
1 0
t+1 y
+y =
1
t−5 ,
y(0) = 1, y 0 (0) = 2
Example 10.5
Is there a solution y(t) to the initial value problem
y 00 + 2y 0 +
1
y = 0, y(1) = 1, y 0 (1) = 2
t−3
such that limt→0+ y(t) = ∞? That is, y(t) has an infinite discontinuity at t = 0.
Solution.
1
Since p(t) = 2, q(t) = t−3
, and t0 = 1, according to Theorem 10.1 the largest
interval for which the solution y(t) is defined is −∞ < t < 3. Since 0 is in that
interval and y(t) is continuous , the limit cannot hold
79
Practice Problems
In Problems 10.1 - 10.6, determine the largest t−interval on which Theorem 10.1
guarantees the existence of a unique solution.
Problem 10.1
y 00 + y 0 + 3ty = tan t, y(π) = 1, y 0 (π) = −1.
Problem 10.2
et y 00 +
1
y
t2 −1
= 4t , y(−2) = 1, y 0 (−2) = 2.
Problem 10.3
ty 00 +
sin 2t 0
y
t2 −9
+ 2y = 0, y(1) = 0, y 0 (1) = 1.
Problem 10.4
ty 00 − (1 + t)y 0 + y = t2 e2t , y(−1) = 0, y 0 (−1) = 1.
Problem 10.5
(sin2 t)y 00 − (2 sin t cos t)y 0 + (cos2 t + 1)y = sin3 t, y( π4 ) = 0, y 0 ( π4 ) =
√
2.
Problem 10.6
t2 y 00 + ty 0 + y = sec (ln t), y( π3 ) = 0, y 0 ( π3 ) = −1.
In Problems 10.7 - 10.8, give an example of an initial value problem of the form
y 00 + p(t)y 0 + q(t)y = 0, y(t0 ) = y0 , y 0 (t0 ) = y00 for which the given t−interval is
the largest on which Theorem 10.1 guarantees a unique solution.
Problem 10.7
−∞ < t < ∞.
8010 SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQU
Problem 10.8
3 < t < ∞.
Problem 10.9
Determine the largest interval in which the initial-value problem
(t − 3)y 00 + ty 0 + (ln |t|)y = 0, y(1) = 0, y 0 (1) = 1
is certain to have a unique solution.
Problem 10.10
Theorem 10.1 tells us that the initial-value problem
y 00 + t2 y = 0, y(0) = y 0 (0) = 0
defines exactly one function y(t). Using only Theorem 10.1, show that this function
has the additional property y(−t) = y(t). That is, y(t) is an even function.
Problem 10.11
State the existence and uniqueness theorem for second order linear differential
equations with initial consitions.
Problem 10.12
Consider the homogeneous second order linear differential equation
ty 00 − y 0 + 4t3 y = 0.
Show that y = cos (t2 ) is a solution to this differential equation.
Problem 10.13
Find the largest interval where
p
16 − t2 y 00 + ln (t + 1)y 0 + (cos t)y = 0, y(0) = 2, y 0 (0) = 0
is guaranteed to have a unique solution.
Problem 10.14
Find the largest interval where
(t + 2)y 00 + ty 0 + cot ty = t2 + 1, y(2) = 11, y 0 (2) = −2
is guaranteed to have a unique solution.
Problem 10.15
Define L(y) = y 00 +p(t)y 0 +q(t)y. Show that L(αy1 +βy2 ) = αL(y1 )+βL(y2 ), where
y1 and y2 are twice differentiable functions. That is, L is a linear transformation.
11 The General Solution of 2nd
Order Linear Homogeneous
Equations
In this section we discuss the structure of the general solution to the homogeneous
second order linear differential equation
y 00 + p(t)y 0 + q(t)y = 0
(11.1)
where p(t) and q(t) are continuous functions for a < t < b.
The first key property of (11.1) is its linear property also known as the principle
of superposition.
Theorem 11.1 (Principle of Superposition)
If y1 and y2 are respective solutions of (11.1) then for any constants c1 and c2 , the
function y = c1 y1 + c2 y2 is also a solution to (11.1).
The function c1 y1 + c2 y2 is called a linear combination of the functions y1 and
y2 .
Example 11.1
Write y = 3 cos 2t + π4 as a linear combination of y1 = cos 2t and y2 = sin 2t.
Solution.
Using the identity
cos (α + β) = cos α cos β − sin α sin β
81
8211 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQUATIONS
we arrive at
π
π
π
3 cos 2t +
=3 cos 2t cos − 3 sin 2t sin
4
4
4
h
h
πi
πi
= 3 cos
cos 2t + −3 sin
sin 2t
4
4
√ !
√ !
3 2
3 2
=
cos 2t + −
sin 2t
2
2
Theorem 11.1 states that if y1 and y2 are two given solutions of (11.1) then one
can build many new solutions by taking a linear combination y = c1 y1 + c2 y2 .
However, this theorem does not say if every solution to (11.1) has to be written
as a linear combination of y1 and y2 . So our next interest is to find out if one can
express every solution of (11.1) as a linear combination of two solutions of (11.1).
If there are such solutions y1 and y2 , we shall say that the set {y1 , y2 } forms a
fundamental set of solutions to (11.1).
It follows that if we know a fundamental set of solutions {y1 , y2 } then we know
the general solution to (11.1) which is given by
y(t) = c1 y1 (t) + c2 y2 (t).
Identifying Fundamental Sets
Given a particular homogeneous differential equation and two solutions of that
differential equation, is there a convenient way for checking whether or not these
two solutions form a fundemental set of solutions? The answer is in the affirmative
according to the following theorem.
Theorem 11.2
Let y1 (t) and y2 (t) be two solutions to the homogeneous second order linear differential equation
y 00 + p(t)y 0 + q(t)y = 0,
a<t<b
where p(t) and q(t) are continuous in a < t < b. If there is a a < t0 < b such that
y1 (t0 ) y2 (t0 ) = y1 (t0 )y20 (t0 ) − y10 (t0 )y2 (t0 ) 6= 0
W (y1 (t0 ), y2 (t0 )) = 0
y1 (t0 ) y20 (t0 ) then {y1 , y2 } is a fundamental set of solutions. We call the function W the Wronskian function.
Proof.
We need to show that if y(t) is a solution to (11.1) then we can write y(t) as a
linear combination of y1 and y2 . That is
y(t) = c1 y1 (t) + c2 y2 (t).
83
So the problem reduces to finding the constants c1 and c2 . These are found by
solving the following linear system of two equations in the unknowns c1 and c2 :
c1 y1 (t0 ) + c2 y2 (t0 ) = y(t0 )
c1 y10 (t0 ) + c2 y20 (t0 ) = y 0 (t0 ).
By the method of elimination we find
c1 =
y(t0 )y20 (t0 ) − y 0 (t0 )y2 (t0 )
W (y1 (t0 ), y2 (t0 ))
c2 =
y 0 (t0 )y1 (t0 ) − y(t0 )y10 (t0 )
.
W (y1 (t0 ), y2 (t0 ))
and
Note that c1 and c2 exist since W (y1 (t0 ), y2 (t0 )) 6= 0
Example 11.2
Consider the differential equation
y 00 + 4y = 0.
(11.2)
(a) Show that y1 (t) = cos 2t and y2 (t) = sin 2t are solutions to (11.2).
(b) Show that {cos 2t, sin 2t} is a fundamental set of solutions.
(c) Write the solution y(t) = 3 cos (2t + π4 ) as a linear combination of y1 and y2 .
Solution.
(a) A simple calculation shows
y100 + 4y1 = −4 cos 2t + 4 cos 2t = 0
y200 + 4y2 = −4 sin 2t + 4 sin 2t = 0.
(b) For any t we have
cos 2t
sin 2t
W (y1 (t), y2 (t)) = −2 sin 2t 2 cos 2t
= 2 cos2 2t + 2 sin2 2t = 2 6= 0.
Thus, {y1 , y2 } is a fundamental set of solutions.
(c) Using the formulas for c1 and c2 with t0 = 0 we find
y(0)y20 (0) − y 0 (0)y2 (0)
W (y1 (0), y2 (0))
√
6 cos π4 cos 0 + 6 sin π4 sin 0
3 2
=
=
2
2
c1 =
8411 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQUATIONS
and
y 0 (0)y1 (0) − y(0)y10 (0)
W (y1 (0), y2 (0))
√
−6 sin π4 cos 0 + 6 cos π4 sin 0
3 2
=
=−
2
2
c2 =
Theorem 11.2 says that if one can find a < t0 < b such that W (y1 (t0 ), y2 (t0 )) 6= 0
then the set {y1 , y2 } is a fundamental set of solutions. The following theorem
shows that the condition W (y1 (t0 ), y2 (t0 )) 6= 0 implies that W (y1 (t), y2 (t)) 6= 0
for all t in the interval (a, b). That is, the theorem tells us that we can choose our
test point t0 on the basis of convenience−any test point t0 will do. That’s why we
chose t0 = 0 in the previous example.
Theorem 11.3 (Abel’s)
Let y1 (t) and y2 (t) be two solutions to the homogeneous second order linear differential equation
y 00 + p(t)y 0 + q(t)y = 0,
a<t<b
where p(t) and q(t) are continuous in a < t < b. If t0 is any point in (a, b) then
W (y1 (t), y2 (t)) = W (y1 (t0 ), y2 (t0 ))e
−
Rt
t0
p(s)ds
.
Thus, if W (y1 (t0 ), y2 (t0 )) 6= 0 then W (y1 (t), y2 (t)) 6= 0 for all a < t < b.
Remark 11.1
It is easy to check that W satisfies the differential equation W 0 + pW = 0. Thus,
Abel’s formula is just the solution to the initial value problem
W 0 + pW = 0, W (y1 (t0 ), y2 (t0 )) = α 6= 0.
In the case y1 and y2 form a fundamental set of solutions then W (y1 (t), y2 (t)) is
never zero in the interval a < t < b as shown in the following theorem.
Theorem 11.4
Suppose that {y1 , y2 } is a fundamental set of solutions to (11.1). Then W (y1 (t), y2 (t)) 6=
0 for all a < t < b.
Combining Theorem 11.2 and Theorem 11.4 we obtain the following corollary
characterizing a fundamental set of solutions.
Corollary 11.1
Let y1 (t) and y2 (t) be two solutions of (11.1). Let W (y1 (t), y2 (t)) denote the
Wronskian of y1 and y2 . Then {y1 , y2 } is a fundamental set of solution if and only
if W (y1 (t), y2 (t)) 6= 0 for all a < t < b.
85
Example 11.3
Consider the initial value problem
1
3
y 00 − y 0 − 2 y = 0, y(1) = 4, y 0 (1) = 8, 0 < t < ∞.
t
t
(a) Show that y1 (t) = t3 and y2 (t) = t−1 are solutions to the differential equation.
(b) Show that {y1 , y2 } is a fundamental set of solutions to the differential equation.
(c) Solve the given initial value problem.
Solution.
(a) By substitution and simple calculation we find
1
3
1
3
y100 − y10 − 2 y1 = 6t − · 3t2 − 2 · t3 = 0
t
t
t
t
1
3
1
3
y200 − y20 − 2 y2 = 2t−3 − · (−t−2 ) − 2 · t−1 = 0.
t
t
t
t
(b) Finding the Wronskian at t0 = 1 we see
1 1 = −4 6= 0.
W (y1 (1), y2 (1)) = 3 −1 Thus, {y1 , y2 } is a fundamental set of solution.
(c) The general solution to the differential equation has the form y(t) = c1 y1 (t) +
c2 y2 (t). The initial conditions yield the following linear system in the unknowns
c1 and c2 .
c1 y1 (1) + c2 y2 (1) =4
c1 y10 (1) + c2 y20 (1) =8
or
c1 + c2 =4
3c1 − c2 =8.
Solving this system by the method of elimination we find c1 = 3 and c2 = 1. Thus,
y(t) = 3t3 + 1t , 0 < t < ∞
Example 11.4
Suppose that y1 (t) = t is a solution to the differential equation
t2 y 00 − (t + 2)ty 0 + (t + 2)y = 0.
Find a second solution y2 .
8611 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQUATIONS
Solution.
Rewriting the given equation in the form
2
2
1
y 00 − ( + 1)y 0 + ( 2 + )y = 0.
t
t
t
Thus, p(t) = −( 2t + 1). But W 0 + pW = 0 so that
2
W 0 − ( + 1)W = 0.
t
Using the method of integrating factor we find
W (t) = Ct2 et .
To find y2 , note that
y2 0
t
=
W (t)
t2
= et . Integrating both sides, we find
Z
y2 (t) = t et dt = tet
where we chose the constant of integration to be 0
87
Practice Problems
In Problems 11.1-11.7, the t−interval of solution is −∞ < t < ∞ unless indicated otherwise.
(a) Determine whether the given functions are solutions to the differential equation.
(b) If both functions are solutions, calculate the Wronskian. Does this calculation
show that the two functions form a fundamental set of solutions?
(c) If the two functions have been shown in (b) to form a fundamental set, construct the general solution and determine the unique solution satisfying the initial
value problem.
Problem 11.1
y 00 − 4y = 0, y1 (t) = e2t , y2 (t) = 2e−2t , y(0) = 1, y 0 (0) = −2.
Problem 11.2
y 00 + y = 0, y1 (t) = sin t cos t, y2 (t) = sin t, y( π2 ) = 1, y 0 ( π2 ) = 1.
Problem 11.3
y 00 − 4y 0 + 4y = 0, y1 (t) = e2t , y2 (t) = te2t , y(0) = 2, y 0 (0) = 0.
Problem 11.4
ty 00 + y 0 = 0, y1 (t) = ln t, y2 (t) = ln 3t, y(3) = 0, y 0 (3) = 3, 0 < t < ∞.
Problem 11.5
t2 y 00 − ty 0 − 3y = 0, y1 (t) = t3 , y2 (t) = −t−1 , y(−1) = 0, y 0 (−1) = −2, t < 0.
Problem 11.6
y 00 = 0, y1 (t) = t + 1, y2 (t) = −t + 2, y(1) = 4, y 0 (1) = −1.
8811 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQUATIONS
Problem 11.7
t
t
4y 00 + 4y 0 + y = 0, y1 (t) = e 2 , y2 (t) = te 2 , y(1) = 1, y 0 (1) = 0.
Problem 11.8
The functions y1 (t) = t and y2 (t) = t ln t form a fundamental set of solutions to
the differential equation
t2 y 00 − ty 0 + y = 0, 0 < t < ∞.
(a) Show that y(t) = 2t + t ln 3t is a solution to the differential equation.
(b) Find c1 and c2 such that y(t) = c1 y1 (t) + c2 y2 (t).
Problem 11.9
The functions y1 (t) = e3t and y2 (t) = e−3t are known to be solutions of y 00 + αy 0 +
βy = 0, where α and β are constants. Determine α and β.
Problem 11.10
The functions y1 (t) = t and y2 (t) = et are known to be solutions of y 00 + p(t)y 0 +
q(t)y = 0.
(a) Determine the functions p(t) and q(t).
(b) On what t−intervals are the functions p(t) and q(t) continuous?
(c) Compute the Wronskian of these two functions. On what t−intervals is the
Wronskian nonzero?
(d) Are the observations in (b) and (c) consistent with Abel’s Theorem?
Problem 11.11
It is known that two solutions of y 00 + ty 0 + 2y = 0 has a Wronskian W (y1 (t), y2 (t))
that satisfies W (y1 (1), y2 (1)) = 4. What is W (y1 (2), y2 (2))?
Problem 11.12
The pair of functions {y1 , y2 } is known to form a fundamental set of solutions of
y 00 + αy 0 + βy = 0, where α and β are constants. One solution is y1 (t) = e2t , and
the Wronskian formed by these two solutions is W (y1 (t), y2 (t)) = e−t . Determine
the constants α and β.
Problem 11.13
2
The Wronskian of a pair of solutions of y 00 + p(t)y 0 + 3y = 0 is W (t) = e−t . What
is the coefficient function p(t)?
89
Problem 11.14
Prove that if y1 and y2 have maxima or minima at the same point in an interval
I, then they cannot be a fundamental set of solutions on that interval.
Problem 11.15
Without solving the equation, find the Wronskian of two solutions of Bessel’s
equation
t2 y 00 + ty 0 + (t2 − µ2 )y = 0.
Problem 11.16
If W (y1 , y2 ) = t2 et and y1 (t) = t then find y2 (t).
Problem 11.17
The functions t2 and 1/t are solutions to a 2nd order, linear homogeneous ODE
on t > 0. Verify whether or not the two solutions form a fundamental solution set.
Problem 11.18
Show that t3 and t4 can’t both be solutions to a differential equation of the form
y 00 + p(t)y 0 + q(t)y = 0 where p and q are continuous functions defined on the real
numbers.
Problem 11.19
Suppose that t2 + 1 is the Wronskian of two solutions to the differential equation
y 00 + p(t)y 0 + q(t)y = 0. Find p(t).
9011 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQUATIONS
12 Second Order Linear
Homogeneous Equations with
Constant Coefficients: Distinct
Characterisitc Roots
In the previous section we discussed the structure of the general solution of a second
order linear homogeneous differential equation. As we saw, the general solution
is a linear combination of two solutions that form a fundamental set of solutions.
In this and the next two sections we discuss methods for finding the fundamental
set of solutions for second order homogeneous equations with constant coefficients,
i.e., equations of the form
ay 00 + by 0 + cy = 0
(12.1)
where a, b and c are constants with a 6= 0.
Notice first that for b = 0 and c 6= 0 the function y 00 is a constant multiple of y.
So it makes sense to look for a function with such property. One such function is
y(t) = ert . Substituting this function into (12.1) leads to
ay 00 + by 0 + cy = ar2 ert + brert + cert = (ar2 + br + c)ert = 0.
Since ert > 0 for all t, the previous equation leads to
ar2 + br + c = 0.
(12.2)
Thus, a function y(t) = ert is a solution to (12.1) when r satisfies equation (12.2).
We call (12.2) the characteristic equation for (12.1) and the polynomial C(r) =
ar2 + br + c is called the characteristic polynomial.
Example 12.1
Solve: y 00 − 5y 0 − 6y = 0.
91
9212 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTANT COEF
Solution.
The characteristic polynomial for this equation is C(r) = r2 −5r−6 = (r−2)(r−3).
Thus, the roots of the characteristic equation are r = 2 and r = 3. Since
2t
e
e3t W (t) = 2t
= e5t 6= 0
2e
3e3t the functions y1 (t) = e2t and y2 (t) = e3t form a fundamental set of solutions.
Hence, the general solution is given by y(t) = c1 e2t + c2 e3t where c1 and c2 are
arbitrary constants
We conclude from the previous example that the two distinct real solutions to
the characteristic equation lead to the general solution. Does this result apply
to any equation (12.1) whose characteristic equation has distinct solutions? The
answer is in the affirmative. To see this, let r1 and r2 be the two distinct solutions
to (12.2). Then
rt
r2 t e1
e
= r2 e(r1 +r2 )t − r1 e(r1 +r2 )t = (r2 − r1 )e(r1 +r2 )t 6= 0
W (t) = r1 er1 t r2 er2 t since both r1 − r2 and e(r1 +r2 )t are not equal to 0. Hence, er1 t and er2 t form a
fundamental set of solutions. As a result, the general solution of (12.1) is given by
y(t) = c1 er1 t + c2 er2 t where c1 and c2 are arbitrary constants.
Example 12.2
Solve the initial value problem
y 00 − y 0 − 6y = 0, y(0) = 1, y 0 (0) = 2.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Solution.
The characteristic polynomial is C(r) = r2 − r − 6 = (r − 3)(r + 2) so that the
characteristic equation r2 − r − 6 = 0 has the solutions r1 = 3 and r2 = −2. The
general solution is then given by
y(t) = c1 e3t + c2 e−2t .
Taking the derivative to obtain
y 0 (t) = 3c1 e3t − 2c2 e−2t .
93
The conditions y(0) = 1 and y 0 (0) = 2 lead to the system
c1 + c2 = 1
3c1 − 2c2 = 2.
Solving this system by the method of elimination we find c1 =
Hence, the unique solution to the initial value problem is
4
5
and c2 =
1
5.
1
y(t) = (4e3t + e−2t ).
5
As t → −∞, e3t → 0 and e−2t → ∞. Thus, y(t) → ∞. Similarly, y(t) → ∞ as
t→∞
Remark 12.1
In this section, we have only considered the case when (12.2) has two distinct real
solutions. In Section 13, we discuss the case of (12.2) having repeated solutions
and in Section 14 we look at the complex solutions
Example 12.3
Consider the initial value problem y 00 + αy 0 + βy = 0, y(0) = 1, y 0 (0) = y00 ,
where α, β, and y00 are constants. It is known that one solution of the differential
equation is y1 (t) = e−3t and that the solution of the initial value problem satisfies
limt→∞ y(t) = 2. Determine the constants α, β, and y00 .
Solution.
Since r = −3 is a solution to the characteristic equation, we have (−3)2 + α(−3) +
β = 0 or −3α + β = −9. Also, since limt→∞ y(t) = 2, the second root for the
characteristic equation must be r = 0. In this case, β = 0 and solving for α we
find α = 3. Hence, y(t) = c1 e−3t + c2 . Since limt→−∞ y(t) = 2, we find c2 = 2.
Since y(0) = 1, we find c1 + 2 = 1 so that c1 = −1. Thus, y(t) = −e−3t + 2 and
y 0 (t) = 3e−3t . Therefore, y00 = y 0 (0) = 3
Example 12.4
Consider the initial value problem y 00 + αy 0 + βy = 0, y(0) = 3, y 0 (0) = 5. The
differential equation has a fundamental set of solutions {y1 , y2 }. It is known that
y1 (t) = e−t and that the Wronskian formed by the two members of the fundamental
set is W (t) = 4e2t .
(a) Determine y2 (t).
(b) Determine the constants α and β.
(c) Solve the initial value problem.
9412 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTANT COEF
Solution.
(a) The second solution is of the form y2 (t) = ert . In this case,
−t
e
ert W (t) = = (r + 1)e(r−1)t .
−e−t rert But W (t) = 4e2t and this leads to r = 3. Hence, y2 (t) = e3t .
(b) Since r = −1 and r = 3 are the roots for the characteristic equation, we have
(r + 1)(r − 3) = 0 or r2 − 2r − 3 = 0. This implies that y 00 − 2y 0 − 3y = 0 so that
α = −2 and β = −3
(c) The initial conditions and y 0 (t) = −c1 e−t +3c2 e3t lead to the system c1 +c2 = 3
and −c1 + 3c2 = 5. Solving this system we find c1 = 1 and c2 = 2. Thus,
y(t) = e−t + 2e3t
Example 12.5
Solve the initial-value problem 4y 00 − y = 0, y(0) = 2, y 0 (0) = β. Then find β so
that the solution approaches zero as t → ∞.
Solution.
The characteristic equation 4r2 − 1 = 0 has roots r = − 21 and r = 12 . Thus,
t
t
y(t) = c1 e− 2 + c2 e 2 .
t
t
The initial conditions and y 0 (t) = − c21 e− 2 + c22 e 2 lead to the system c1 + c2 = 2
and c1 − c2 = −2β. Solving this system we find c1 = 1 − β and c2 = 1 + β. Thus,
t
t
y(t) = (1 − β)e− 2 + (1 + β)e 2 .
Since limt→∞ y(t) = 0 and limt→∞ et = ∞, we must have 1+β = 0. Thus, β = −1
Example 12.6
Show that if λ is a root of aλ3 + bλ2 + cλ + d = 0, then eλt is a solution of
ay 000 + by 00 + cy 0 + dy = 0.
Solution.
We have
ay 000 + by 00 + cy 0 + dy = aλ3 eλt + bλ2 eλt + cλeλt + deλt
= (aλ3 + bλ2 + cλ + d)eλt = 0
95
Practice Problems
Problem 12.1
Solve the initial value problem
y 00 + y 0 − 2y = 0, y(0) = 3, y 0 (0) = −3.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.2
Solve the initial value problem
y 00 − 4y 0 + 3y = 0, y(0) = −1, y 0 (0) = 1.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.3
Solve the initial value problem
y 00 − y = 0, y(0) = 1, y 0 (0) = −1.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.4
Solve the initial value problem
y 00 + 5y 0 + 6y = 0, y(0) = 1, y 0 (0) = −1.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.5
Solve the initial value problem
y 00 − 4y = 0, y(3) = 0, y 0 (3) = 0.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.6
Solve the initial value problem
2y 00 − 3y 0 = 0, y(−2) = 3, y 0 (−2) = 0.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
9612 SECOND ORDER LINEAR HOMOGENEOUS EQUATIONS WITH CONSTANT COEF
Problem 12.7
Solve the initial value problem
y 00 + 4y 0 + 2y = 0, y(0) = 0, y 0 (0) = 4.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.8
Solve the initial value problem
2y 00 − y = 0, y(0) = −2, y 0 (0) =
√
2.
Describe the behavior of the solution y(t) as t → −∞ and t → ∞.
Problem 12.9
Obtain the general solution to the differential equation y 000 − 5y 00 + 6y 0 = 0.
Problem 12.10
A particle of mass m moves along the x−axis and is acted upon by a drag force
proportional to its velocity. The drag constant is denoted by k. If x(t) represents
the particle position at time t, Newton’s law of motion leads to the differential
equation mx00 (t) = −kx0 (t).
(a) Obtain the general solution to this second order linear differential equation.
(b) Solve the initial value problem if x(0) = x0 and x0 (0) = v0 .
(c) What is limt→∞ x(t)?
Problem 12.11
Find a homogeneous second-order linear ordinary differential equation whose general solution is y(t) = c1 e2t + c2 e−t .
Problem 12.12
Find the general solution of the differential equation y 00 − 3y 0 − 4y = 0.
Problem 12.13
Find the general solution of the differential equation y 00 + 4y 0 − 5y = 0.
Problem 12.14
Find the general solution of the differential equation −3y 00 + 2y 0 + y = 0.
Problem 12.15
Solve the initial-value problem: y 00 + 3y 0 − 4y = 0, y(0) = −1, y 0 (0) = 1.
Problem 12.16
Solve the initial-value problem: 2y 00 + 5y 0 − 3y = 0, y(0) = 2, y 0 (0) = 1.
13 Characteristic Equations
with Repeated Roots
In this section we consider the question of the characteristic equation having a
repeated real solution. This occurs when b2 − 4ac = 0. The two equal roots are
given by
b
r1 = r2 = − .
2a
The computation based on the trial form y(t) = ert yields only one solution,
namely
b
y1 (t) = e− 2a t .
Since a fundamental set of solutions consists of two functions having a nonzero
Wronskian, there must be another solution having a different functional form.
The second solution follows from the following theorem.
Theorem 13.1
Suppose that y1 (t) is a nontrivial solution to the differential equation
y 00 + p(t)y 0 + q(t)y = 0.
Then any solution y2 (t) can be written in the form
Z − R p(t)dt !
e
y2 (t) = C
dt y1 (t) + C 0 y1 (t)
y12 (t)
(13.1)
(13.2)
where C and C 0 are arbitrary constants.
Proof.
First, recall that the Wronskian W (t) of any two solutions
R to (13.1) satisfies the
differential equation W 0 + p(t)W = 0 so that W (t) = Ce− p(t)dt . If y2 is a solution
to (13.1) then
R
0
W (t)
e− p(t)dt
y2
= 2
=C
.
y1
y1 (t)
y12 (t)
97
98
13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS
Integrating this last equation we find
Z
y2 (t) = C
!
R
e− p(t)dt
dt y1 (t) + C 0 y1 (t)
y12 (t)
where C and C 0 are arbitrary constants
Notice that the term C 0 y1 (t) is simply a constant multiple of y1 (t). Since the
general solution of the differential equation (13.1) contains y1 (t) multiplied by an
arbitrary constant, we lose no generality by setting C 0 = 0. We can likewise take
C = 1 since y2 (t) will also be multiplied by an arbitrary constant in the general
solution. With these simplification the second solution is
Z
y2 (t) =
!
R
e− p(t)dt
dt y1 (t).
y12 (t)
Now, for the equation
ay 00 + by 0 + cy = 0
(13.3)
b
we have p(t) = ab . If y1 (t) = e− 2a t then
Z
y2 (t) =
!
b
e− a t
e
− ab t
b
b
dt e− 2a t = te− 2a t .
Hence, the general solution to (13.3) is given by
b
b
y(t) = c1 e− 2a t + c2 te− 2a t .
Example 13.1
Solve the initial value problem: y 00 + 2y 0 + y = 0, y(0) = 1, y 0 (0) = −1.
Solution.
The characteristic equation r2 + 2r + 1 = 0 has a repeated root: r1 = r2 = −1.
Thus, the general solution is given by
y(t) = c1 e−t + c2 te−t .
The two conditions y(0) = 1 and y 0 (0) = −1 lead to c2 = 0 and c1 = 1. Hence, the
unique solution is y(t) = e−t
99
Example 13.2
Consider the differential equation
t2 y 00 + 2ty 0 − 2y = 0, 0 < t < ∞.
Find the general solution given that y1 (t) = t is a solution to the differential
equation.
Solution.
Since t > 0, we can rewrite the given equation in the form
2
2
y 00 + y 0 − 2 y = 0.
t
t
In this case, p(t) =
2
t
and
Z
y2 (t) =
e−
R
t2
2
dt
t
!
Z
dt t =
1
1
dt t = − 2 .
4
t
3t
Hence, the general solution is y(t) = c1 t + c2 t−2
Example 13.3
The graph of a solution y(t) of the differential equation 4y 00 + 4y 0 + y = 0 passes
1
through the points (1, e− 2 ) and (2, 0). Determine y(0) and y 0 (0).
Solution.
The characteristic equation 4r2 + 4r + 1 = 0 has the roots r1 = r2 = − 21 so that
the general solution is
t
t
y(t) = c1 e− 2 + c2 te− 2 .
1
Since y(2) = 0, we find c1 + 2c2 = 0. Since y(1) = e− 2 , we have c1 + c2 = 1.
Solving the system of two equations, we find c1 = 2 and c2 = −1. Hence,
t
t
y(t) = 2e− 2 − te− 2 .
t
t
Now, y(0) = 2. Also, replacing t = 0 in y 0 (t) = −2e− 2 + 2t e− 2 to obtain y 0 (0) =
−2
Example 13.4
Find a homogeneous second order linear differential equation whose general solution is given by y(t) = c1 e−3t + c2 te−3t .
100
13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS
Solution.
The characteristic equation has the double roots r1 = r2 = −3 so that r2 +6r +9 =
0. Hence, the differential equation is
y 00 + 6y 0 + 9y = 0
Example 13.5
Show that if λ is a double root of at3 + bt2 + ct + d = 0, then teλt is also a solution
of ay 000 + by 00 + cy 0 + dy = 0.
Solution.
Since λ is a double root, we have aλ3 + bλ2 + cλ + d = 0 and 3aλ2 + 2bλ + c = 0.
Let y(t) = teλt . Then y 0 = eλt + λteλt , y 00 = 2λeλt + λ2 teλt , y 000 = 3λ2 eλt + λ3 teλt .
Substituting into the equation we find
ay 000 + by 00 + cy 0 + dy =[(aλ3 + bλ2 + cλ + d)t + (3aλ2 + 2bλ + c)]eλt
=0
101
Practice Problems
In Problems 13.1 - 13.5 answer the following questions.
(a) Obtain the general solution of the differential equation.
(b) Impose the initial conditions to obtain the unique solution of the initial value
problem.
(c) Describe the behavior of the solution as t → −∞ and t → ∞.
Problem 13.1
5
9y 00 − 6y 0 + y = 0, y(3) = −2, y 0 (3) = − .
3
Problem 13.2
3
25y 00 + 20y 0 + 4y = 0, y(5) = 4e−2 , y 0 (5) = − e−2 .
5
Problem 13.3
y 00 − 4y 0 + 4y = 0, y(1) = −4, y 0 (1) = 0.
Problem 13.4
√
y 00 + 2 2y 0 + 2y = 0, y(0) = 1, y 0 (0) = 0.
Problem 13.5
√
√
3y 00 + 2 3y 0 + y = 0, y(0) = 2 3, y 0 (0) = 3.
Problem 13.6
Find the general solution of y 00 − 6y 0 + 9y = 0.
Problem 13.7
Find the general solution of 4y 00 − 4y 0 + y = 0.
Problem 13.8
Solve the initial-value problem: y 00 + y 0 +
y
4
= 0, y(0) = 2, y 0 (0) = 0.
102
13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS
14 Characteristic Equations
with Complex Roots
In this section, we solve the linear second order homogeneous differential equation
with constant coefficients
ay 00 + by 0 + cy = 0, a 6= 0
(14.1)
when the roots of the characteristic equation
ar2 + br + c = 0
(14.2)
are complex numbers. This occurs when b2 − 4ac < 0. In this case, the complex
roots of equation (14.2) are given by
√
−b ± i 4ac − b2
r1,2 =
2a
√
where i = −1. We will write
r1,2 = α ± iβ
b
where α = − 2a
and β =
functions
√
4ac−b2
.
2a
Like before, we would like to conclude that the
e(α+iβ)x and e(α−iβ)x
are solutions to (14.1). These are complex solutions, we would like to have real
solutions to the original real differential equation. This requires the use of the
so-called the complex exponential function which we introduce and discuss
some of its properties.
For any complex number z = α + iβ we define the Euler’s function
ez = eα (cos β + i sin β).
103
104
14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS
The exponential function satisfies the usual laws of exponentials such as
ez ew = ez+w .
To see this, we let z = α1 + iβ1 and w = α2 + iβ2 . Then
ez ew =eα1 (cos β1 + i sin β1 )eα2 (cos β2 + i sin β2 )
=eα1 +α2 [(cos β1 cos β2 − sin β1 sin β2 ) + i(sin β1 cos β2 + cos β1 sin β2 )]
=eα1 +α2 [cos (β1 + β2 ) + i sin (β1 + β2 )]
=ez+w .
From the above rule we can write
(ez )n = ez · ez · · · ez = ez+z+···+z = enz
where n is a positive integer.
It follows from the above discussion that the complex solutions to the differential equation are linear combinations of eαt cos βt and eαt sin βt. Now letting
y1 (t) = eαt cos βt and y2 (t) = eαt sin βt we find
ay100 + by10 + cy1 =a(α2 eαt cos βt − β 2 eαt cos βt − 2αβeαt sin βt)
+b(αeαt cos βt − βeαt sin βt) + ceαt cos βt
=eαt cos βt(a(α2 − β 2 ) + bα + c) − eαt sin βt(2aαβ + bβ)
2
b
4ac − b2
−b
αt
=e cos βt a
−
+b
+c
4a2
4a2
2a
!
!
√
√
b 4ac − b2
b 4ac − b2
αt
+
= 0.
−e sin βt 2a −
2a
2a
2a
Thus, y1 (t) = eαt cos βt is a solution to equation (14.1). Similarly, we show that
y2 (t) = eαt sin βt is a solution to equation (14.1). Moreover,
eαt cos βt
eαt sin βt
= βe2αt 6= 0.
W (t) = αt
αt
αt
αt
αe cos βt − βe sin βt αe sin βt + βe cos βt Hence, {y1 , y2 } is a fundamental set of solutions to equation (14.1) so that the
general solution is given by
y(t) = eαt (c1 cos βt + c2 sin βt)
where c1 and c2 are real numbers.
105
Example 14.1
Solve: y 00 + 2y 0 + 5y = 0.
Solution.
The characteristic equation r2 + 2r + 5 = 0 has complex roots r1,2 = −1 ± 2i. The
general solution is
y(t) = e−t (c1 cos 2t + c2 sin 2t)
Example 14.2
Solve the initial value problem
y 00 − 10y 0 + 29y = 0, y(0) = 1, y 0 (0) = 3.
Solution.
The characteristic equation r2 − 10r + 29 = 0 has the complex roots r1,2 = 5 ± 2i.
Thus, the general solution is given by the expression
y(t) = e5t (c1 cos 2t + c2 sin 2t).
Finding y 0 we obtain
y 0 (t) = e5t [(5c1 + 2c2 ) cos 2t + (5c2 − 2c1 ) sin 2t].
The initial conditions yield c1 = 1 and c2 = −1. Thus, the unique solution to the
initial value problem is
y(t) = e5t (cos 2t − sin 2t)
Next, we consider the question of representing the general solution y(t) = eαt (c1 cos βt+
c2 sin βt) in the form y(t) = Keαt cos (βt − δ), where 0 ≤ δ < 2π. For this, we let
P (c1 , c2 ) be a coordinate point in the plane and let δ be the angle between the
−−→
t-axis and ray OP . See Figure 14.1. Then
cos δ =
c1
K
and sin δ =
c2
K
p
c21 + c22 . Then in terms of K and δ we can write
c
c2
1
c1 cos βt + c2 sin βt =K
cos βt +
sin βt
K
K
=K(cos δ cos βt + sin δ sin βt) = K cos (βt − δ).
where K =
It follows that
y(t) = Keαt cos (βt − δ).
106
14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS
Figure 14.1
We call Keαt the amplitude of the oscillations. This means that the graph of
y(t) is bounded by the graphs of ±Keαt . The angle βδ is the phase shift. The
term “phase shift” reflects the fact that we obtain the graph of cos (βt − δ) =
cos β t − βδ by shifting the graph of cos βt to the right by an amount t = βδ .
Example 14.3
Put the solution of the initial value problem
y 00 − 2y 0 + 17y = 0, y(0) = −4, y 0 (0) = 8
in the form y(t) = Keαt cos (βt − δ).
Solution.
The characteristic equation r2 − 2r + 17 = 0 has the complex roots r1,3 = 1 ± 4i.
Thus, the general solution to the differential equation is
y(t) = et (c1 cos 4t + c2 sin 4t).
Since y(0) = −4, we find c1 = −4. Also y 0 (t) = et (c1 cos 4t+c2 sin 4t)+et (4c2 cos 4t−
4c1 sin 4t) and y 0 (0) = 8 imply 8 = −4 + 4c2 so that c2 = 3. Thus, the unique solution to the initial value problem is
y(t) = et (3 sin 4t − 4 cos 4t).
Now, K =
√
9 + 16 =
√
25 = 5. Thus, tan δ = − 34 so that δ = arctan (− 43 ). Hence,
3
y(t) = 5e cos 4t + arctan
.
4
t
107
The graph of y(t) together with the envelope containing it is shown in Figure 14.2
Figure 14.2
108
14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS
Practice Problems
Problem 14.1
For any z = α + iβ we define the conjugate of z to be the complex number
1
(z − z).
z = α − iβ. Show that α = 12 (z + z) and β = 2i
Problem 14.2
Write each of the complex numbers in the form α + iβ, where α and β are real
numbers.
π
1. 2ei 3 .
3π
2. (2√− i)ei 2 .
π
3. ( 2ei 6 )4 .
Problem 14.3
Write each function in the form eαt (A cos βt + iB sin βt), where α, β, A, and B are
real numbers.
√
1. 2ei 2t .
2. − 12 e2t+i(t+π) .
√
3. ( 3e(1+i)t )3 .
In Problems 14.4 - 14.8
(a) Determine the roots of the characteristic equation.
(b) Obtain the general solution as a linear combination of real-valued solutions.
(c) Impose the initial conditions and solve the initial value problem.
Problem 14.4
y 00 + 2y 0 + 2y = 0, y(0) = 3, y 0 (0) = −1.
Problem 14.5
2y 00 − 2y 0 + y = 0, y(−π) = 1, y 0 (−π) = −1.
Problem 14.6
π
1
π
y 00 + 4y 0 + 5y = 0, y( ) = , y 0 ( ) = −2.
2
2
2
Problem 14.7
y 00 + 4π 2 y = 0, y(1) = 2, y 0 (1) = 1.
109
Problem 14.8
9y 00 + π 2 y = 0, y(3) = 2, y 0 (3) = −π.
In Problems 14.9 - 14.10, the function y(t) is a solution of the initial value problem
y 00 +ay 0 +by = 0, y(t0 ) = y0 , y 0 (t0 ) = y00 , where the point t0 is specified. Determine
the constants a, b, y0 , and y00 .
Problem 14.9
y(t) = 2 sin 2t + cos 2t, t0 =
π
.
4
Problem 14.10
π
π
y(t) = et− 6 cos 2t − et− 6 sin 2t, t0 =
π
.
6
In Problems 14.11 - 14.13, rewrite the function y(t) in the form y(t) = Keαt cos (βt − δ),
where 0 ≤ δ < 2π. Use this representation to sketch a graph of the given function,
on a domain sufficiently large to display its main features.
Problem 14.11
y(t) = sin t + cos t.
Problem 14.12√
y(t) = et cos t + 3et sin t.
Problem 14.13
y(t) = e−2t cos 2t − e−2t sin 2t.
Problem 14.14
Consider the differential equation y 00 + ay 0 + 9y = 0, where a is a real number.
Suppose that we know the Wronskian of a fundamental set of solutions of this
differential equation is constant: W (t) = 1 for all real numbers t. Find the general
solution of this differential equation.
Problem 14.15
Rewrite y(t) = 2 cos 7t − 11 sin 7t in phase-angle form. Give the exact function (so
your answer will involve the inverse tangent function.)
110
14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS
Problem 14.16
Find a homogeneous linear ordinary differential equation whose general solution
is y(t) = c1 e2t cos (3t) + c2 e2t sin (3t).
Problem 14.17
Rewrite y(t) = 5e(5−2i)t − 3e(5+2i)t , without complex exponents, using sines and
cosines. What ODE of the form ay 00 + by 0 + cy = 0, has y as a solution?
Problem 14.18
Consider the function y(t) = 3 cos 2t − 4 sin 2t. Find a second order linear IVP that
y satisfies.
Problem 14.19
An equation of the form
t2 y 00 + αty 0 + βy = 0, t > 0
where α and β are real constants is called an Euler equation. Show that the
substitution x = ln t transforms Euler equation into an equation with constant
dy dx
coefficients. Hint: dy
dt = dx dt .
Problem 14.20
Use the result of the previous problem to solve the differential equation t2 y 00 +
ty 0 + y = 0.
15 Series Solutions of
Differential Equations
We know how to find the general solution of a second order linear differential
equation with constant coefficients. In this section, we discuss finding the general
solution to the homogeneous equation
y 00 + p(t)y 0 + q(t)y = 0
(15.1)
where p(t) and q(t) are not necessarily constants.
Let t = t0 . We are interested in answering the question: When is it possible to
represent the general solution of (15.1) in terms of a power series that converge in
some open interval centered at t0 ?
In order to answer this question, we need to introduce the following definitions:
We say that a function f (t) is analytic at t = t0 if f (t) can be expressed as a
Taylor series
∞
X
f (n) (t0 )
f (t) =
(t − t0 )n .
n!
n=0
Examples of analytic functions are polynomial functions and rational functions.
We say that t = t0 is an ordinary point when both p(t) and q(t) are analytic at
t0 . If p(t) and/or q(t) is not analytic at t0 , we say that t0 is a singular point.
Example 15.1
Consider the differential equation
5
(1 − t2 )y 00 + (tan t)y 0 + t 3 y = 0
in the open interval −2 < t < 2. Classify each point in this interval as an ordinary
point or a singular point.
Solution.
We first rewrite the equation in the form (15.1),
5
y 00 +
tan t 0
t3
y
+
y = 0.
1 − t2
1 − t2
111
112
15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Therefore,
p(t) =
tan t
1−t2
5
and q(t) =
t3
.
1−t2
The function p(t) is not analytic at t = ±1 (where the denominator vanishes) and
at t = ± π2 (where the tangent has a vertical asymptotes). The function q(t) is
not analytic at t = ±1 and at t = 0 (q(t) does not have a second derivative at 0.)
Thus, in the interval −2 < t < 2, the five points t = 0, ±1, ± π2 are singular points
and all other points are ordinary points
The following theorem answers our earlier question about a series solution to the
equation (15.1).
Theorem 15.1
Let p(t) and q(t) be analytic at t0 and let R be the smaller of the two radii of
convergence1 of their respective Taylor series representations. The general solution
to (15.1) can be expressed as
y(t) =
∞
X
an (t − t0 )n = c1 y1 (t) + c2 y2 (t),
n=0
where c1 and c2 are arbitrary constants. The functions y1 (t) and y2 (t) form a
fundamental set of solutions, analytic in the interval |t − t0 | < R.
Example 15.2
The point t0 = 0 is an ordinary point of the differential equation
y 00 + (t3 + 1)y 0 − ty = 0.
Determine the value of R as stated in the previous theorem.
Solution.
Since p(t) and q(t) are polynomial functions, we have Rp = Rq = ∞. Hence,
R=∞
The Method of Solution
We assume that t0 is an ordinary P
point of the equation (15.1), so it has a series
n
solution near t0 of the form y(t) = ∞
n=0 an (t − t0 ) that converges for |t − t0 | < R.
1
The radius of convergence of a function f (t), about t0 , is the distance between t0 and
the nearest singularity of f (t).
113
We want to determine the coefficients a0 , a1 , a2 , · · · in the series solution.
Let’s differentiate the series termwise twice:
y(t) =a0 + a1 (t − t0 ) + a2 (t − t0 )2 + · · ·
y 0 (t) =a1 + 2a2 (t − t0 ) + 3a3 (t − t0 )2 + · · ·
y 00 (t) =2a2 + 6a3 (t − t0 ) + 12a4 (t − t0 )2 + · · ·
Substituting in the differential equation (15.1), we get
K0 + K1 (t − t0 ) + K2 (t − t0 )2 + · · · = 0
(15.2)
where Kn is function of certain coefficients ai .
In order that series (15.2) be valid for all t in the interval |t − t0 | < R, we must
have
Kn = 0, n = 0, 1, 2, · · · .
This leads to a set of conditions that the coefficients an must satisfy.
Example 15.3
Find the series solution of the differential equation near the ordinary point 0.
(t2 − 4)y 00 + 3ty 0 + y = 0.
Solution.
The singular points of the given differential equation are −2 and 2. Thus, we will
find a series solution of the differential equation in the interval |t| < 2. The solution
is of the form
∞
X
y(t) =
an t n .
n=0
Differentiating this series twice we find
P∞
P
n−2 .
n−1 and y 00 (t) =
y 0 (t) = ∞
n=2 n(n − 1)an t
n=1 nan t
Substituting these series in the given equation we find
t2
∞
X
n(n − 1)an tn−2 − 4
n=2
∞
X
n(n − 1)an tn−2 + 3t
n=2
∞
X
n(n − 1)an tn −
n=2
∞
X
∞
X
∞
X
n=1
∞
X
4n(n − 1)an tn−2 +
n=2
n(n − 1)an tn −
n=0
∞
X
n=0
∞
X
n=0
nan tn−1 +
4(n + 2)(n + 1)an+2 tn +
n=1
∞
X
n=0
3nan tn +
3nan tn +
∞
X
n=0
∞
X
n=0
∞
X
an tn =0
an tn =0
an tn =0
n=0
[n(n − 1)an − 4(n + 1)(n + 2)an+2 + 3nan + an ] tn =0.
114
15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Equating all the coefficients to zero we find
(n + 1)2 an − 4(n + 1)(n + 2)an+2 = 0
and obtain the recurrence formula,
an+2 =
(n + 1)an
, n ≥ 0.
4(n + 2)
From this formula, we find
a0
8
a1
=
6
3a0
=
128
a1
=
30
5a0
=
1024
a1
=
140
..
.
a2 =
a3
a4
a5
a6
a7
..
.
Notice that each even coefficient is expressed in terms of a0 and each odd coefficient
is expressed in terms of a1 . The general solution is:
1 2
3 4
5 6
1 3
1 5
1 7
y(t) = a0 1 + t +
t +
t + · · · +a1 t + t + t +
t + ···
8
128
1024
6
30
140
115
Practice Problems
Problem 15.1
Identify the singular and ordinary points of the differential equation
y 00 + ty 0 + (t2 + 2)y = 0.
Problem 15.2
Identify the singular and ordinary points of the differential equation
1
(t2 − 1)y 00 + ty 0 + y = 0.
t
Problem 15.3
Identify the singular and ordinary points of the differential equation
(t2 − 4)y 00 + 3ty 0 + y = 0.
Problem 15.4
The point t0 = 0 is an ordinary point of the differential equation
(t2 − 4)y 00 + 3ty 0 + y = 0.
Determine a value of R as stated in Theorem 15.1.
Problem 15.5
Find the series solution near the ordinary point 0 to the differential equation
y 00 + ty 0 + (t2 + 2)y = 0.
Problem 15.6
Find the series solution near the ordinary point 0 to the differential equation
y 00 − ty 0 − t2 y = 0.
Problem 15.7
Find the series solution near the ordinary point 0 to the differential equation
(t2 + 1)y 00 − y 0 + y = 0.
Problem 15.8
Write the following as a single power series:
X
X
nan (t − 2)n+1 +
n2 an (t − 2)n .
n=0
n=2
116
15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS
Problem 15.9
Find the series solution near the ordinary point 0 to the differential equation
y 00 + ty 0 + y = 0.
Problem 15.10
Find the series solution near the ordinary point 0 to the differential equation
(1 − t)2 y 00 − 2y = 0.
16 The Structure of the
General Solution of Linear
Nonhomogeneous Equations
In this section, we consider the question of finding the general solution to the
differential equation
y 00 + p(t)y 0 + q(t)y = g(t)
(16.1)
where p(t), q(t) and g(t) are continuous functions for a < t < b. The following
theorem provides the structure of the general solution to equation (16.1).
Theorem 16.1
Let {y1 (t), y2 (t)} be a fundamental set of solutions to the homogeneous equation
y 00 + p(t)y 0 + q(t)y = 0 and yp (t) be a particular solution of the nonhomogeneous
equation
y 00 + p(t)y 0 + q(t)y = g(t), a < t < b.
The general solution of the nonhomogeneous equation is given by
y(t) = yp (t) + c1 y1 (t) + c2 y2 (t)
for constants c1 and c2 .
Proof.
Let y(t) be any solution to equation (16.1). Because yp (t) is also a solution then
(y − yp )00 + p(t)(y − yp )0 + q(t)(y − yp ) =(y 00 + p(t)y 0 + q(t)y)
−(yp00 + p(t)yp0 + q(t)yp )
=g(t) − g(t) = 0.
117
11816 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMOGENEO
Therefore, y − yp is a solution to the homogeneous equation. But {y1 , y2 } is
a fundamental set of solutions to the homogeneous equation so that there exist
unique constants c1 and c2 such that y(t) − yp (t) = c1 y1 (t) + c2 y2 (t). Hence,
y(t) = yp (t) + c1 y1 (t) + c2 y2 (t)
It follows from the above theorem that finding the general solution to nonhomogeneous equations consists of three steps:
1. Find the general solution of the associated homogeneous equation y 00 + p(t)y 0 +
q(t)y = 0.
2. Find a single solution of the original equation y 00 + p(t)y 0 + q(t)y = g(t).
3. Add together the solutions found in steps 1 and 2.
Example 16.1
Use the fact that yp (t) = 3t − 1 to find the unique solution to the initial value
problem
y 00 − 2y 0 − 3y = −9t − 3, y(0) = 1, y 0 (0) = 3.
Solution.
Because we are given yp then we need to find the general solution of the homogeneous equation y 00 − 2y 0 − 3y = 0. The associated characteristic equation
r2 − 2r − 3 = 0 has roots r1 = −1 and r2 = 3. Hence, the general solution to the
differential equation is
y(t) = c1 e−t + c2 e3t + 3t − 1.
The derivative of this function is given by y 0 (t) = −c1 e−t +3c2 e3t +3. The condition
y(0) = 1 leads to c1 + c2 = 2. The condition y 0 (0) = 3 leads to −c1 + 3c2 = 0.
Solving for c1 and c2 we find c1 = 23 and c2 = 12 . The unique solution is given by
3
1
y(t) = e−t + e3t + 3t − 1
2
2
Note that y1 (t) = 3t − 1 and y2 (t) = e3t + 3t − 1 both are particular solutions to
the given differential equation. However, the sum u(t) = y1 (t) + y2 (t) = e3t + 6t − 2
is not a solution since
u00 − 2u0 − 3u = −18t − 6 6= −9t − 3.
This shows that the superposition of solutions is valid only for homogeneous equations and not true in general for nonhomogeneous equations. However, we can
have a property of superposition of nonhomogeneous if one is adding two solutions
of two different nonhomogeneous equations. More precisely, we have
119
Theorem 16.2
Let y1 (t) be a solution of y 00 + p(t)y 0 + q(t)y = g1 (t) and y2 (t) a solution of
y 00 + p(t)y 0 + q(t)y = g2 (t). Then for any constants c1 and c2 the function Y (t) =
c1 y1 (t) + c2 y2 (t) is a solution of the equation
y 00 + p(t)y 0 + q(t)y = c1 g1 (t) + c2 g2 (t).
Proof.
We have
Y 00 + p(t)Y 0 + q(t)Y =c1 y100 + c2 y200 + p(t)c1 y10 + p(t)c2 y20 + q(t)c1 y1 + q(t)c2 y2
=c1 (y100 + p(t)y10 + q(t)y1 ) + c2 (y200 + p(t)y20 + q(t)y2 )
=c1 g1 (t) + c2 g2 (t)
Example 16.2
The functions u1 (t) and u2 (t) are solutions to the following differential equations
u001 + p(t)u01 + q(t)u1 =2e−t − t − 1
u002 + p(t)u02 + q(t)u2 =3t
Use the functions u1 and u2 to construct a particular solution of the differential
equation
u00 + p(t)u0 + q(t)u = 4e−t − 2.
Solution.
The right-hand side of the given equation can be written as 4e−t − 2 = 2(2e−t −
t − 1) + 23 (3t) so that by the previous theorem, the function u(t) = 2u1 (t) + 32 u2 (t)
is the required particular solution
Example 16.3
Consider the IVP
4
y 00 − y 0 − 2y = 4e−t , y(0) = 0, y 0 (0) = 0, yp (t) = − te−t .
3
(a) Verify that the given function, yp (t), is a particular solution of the differential
equation.
(b) Determine the general solution,yh , of the homogeneous equation.
(c) Find the general solution to the differential equation and impose the initial
conditions to obtain the unique solution of the initial value problem.
12016 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMOGENEO
Solution.
(a) yp0 = − 34 e−t + 43 te−t , yp00 = 83 e−t − 34 te−t .
8
4
4
4
8
yp00 − yp0 − 2yp = e−t − te−t + e−t − te−t + te−t
3
3
3
3
3
−t
=4e
(b) The associated characteristic equation r2 − r − 2 = 0 has roots r1 = −1 and
r2 = 2. Hence, the general solution to the homogeneous differential equation is
yh (t) = c1 e−t + c2 e2t
(c) The general solution to the differential equation is y(t) = c1 e−t + c2 e2t − 34 te−t .
The derivative of this function is given by y 0 (t) = −c1 e−t + 2c2 e2t − 34 e−t + 43 te−t .
The condition y(0) = 0 leads to c1 + c2 = 0. The condition y 0 (0) = 0 leads to
−c1 + 2c2 = 34 . Solving for c1 and c2 we find c1 = − 49 and c2 = 94 . The unique
solution is given by
4
y(t) = (e2t − e−t − 3te−t )
9
Example 16.4
Determine the function g(t) if
y 00 − 2y 0 − 3y = g(t), yp (t) = 3e5t .
Solution.
We have yp0 = 15e5t and yp00 = 75e5t . Thus,
g(t) =yp00 − 2yp0 − 3yp = 75e5t − 30e5t − 9e5t = 36e5t
Example 16.5
The general solution of the nonhomogeneous differential equation y 00 + αy 0 + βy =
g(t) is given below, where c1 and c2 are arbitrary constants. Determine the constants α and β and the function g(t).
y(t) = c1 et + c2 e2t + 2e−2t .
Solution.
From the given general solution we see that the roots of the characteristic equation
are r1 = 1 and r2 = 2. Thus, the characteristic equation is (r − 1)(r − 2) =
r2 − 3r + 2 = 0. The associated differential equation is y 00 − 3y 0 + 2y = 0. Hence,
α = −3 and β = 2. Now, letting yp (t) = 2e−t , we have
g(t) =yp00 − 3yp0 + 2yp
=8e−2t + 12e−2t + 4e−2t = 24e−2t
121
Practice Problems
In Problems 16.1- 16.6, answer the following three question.
(a) Verify that the given function, yp (t), is a particular solution of the differential
equations.
(b) Determine the general solution,yh , of the homogeneous equation.
(c) Find the general solution to the differential equation and impose the initial
conditions to obtain the unique solution of the initial value problem.
Problem 16.1
1
y 00 − 2y 0 − 3y = e2t , y(0) = 1, y 0 (0) = 0, yp (t) = − e2t .
3
Problem 16.2
y 00 − y 0 − 2y = 10, y(−1) = 0, y 0 (−1) = 1, yp (t) = −5.
Problem 16.3
y 00 + y 0 = 2e−t , y(0) = 2, y 0 (0) = 2, yp (t) = −2te−t .
Problem 16.4
y 00 + 4y = 10et−π , y(π) = 2, y 0 (π) = 0, yp (t) = 2et−π .
Problem 16.5
π
π
y 00 − 2y 0 + 2y = 5 sin t, y( ) = 1, y 0 ( ) = 0, yp (t) = 2 cos t + sin t.
2
2
Problem 16.6
y 00 − 2y 0 + y = t2 + 4 + 2 sin t, y(0) = 1, y 0 (0) = 3, yp (t) = t2 + 4t + 10 + cos t.
The functions u1 , u2 , and u3 are solutions to the following differential equations
u00 + p(t)u0 + q(t)u =2et + 1
u00 + p(t)u0 + q(t)u =2e−t − t − 1
u00 + p(t)u0 + q(t)u =3t.
In Problems 16.7 - 16.8, use the functions u1 , u2 and u3 to construct a particular
solution of the differential equation.
12216 THE STRUCTURE OF THE GENERAL SOLUTION OF LINEAR NONHOMOGENEO
Problem 16.7
1
u00 + p(t)u0 + q(t)u = et + 2t + .
2
Problem 16.8
u00 + p(t)u0 + q(t)u =
et + e−t
.
2
In Problems 15.9 - 15.11, determine the function g(t).
Problem 16.9
y 00 − 2y 0 = g(t), yp (t) = 3t +
√
t, t > 0.
Problem 16.10
y 00 + y 0 = g(t), yp (t) = ln (1 + t), t > −1.
Problem 16.11
y 00 + 2y 0 + y = g(t), yp (t) = t − 2.
In Problems 16.12 - 16.13, the general solution of the nonhomogeneous differential
equation y 00 + αy 0 + βy = g(t) is given, where c1 and c2 are arbitrary constants.
Determine the constants α and β and the function g(t).
Problem 16.12
y(t) = c1 et + c2 tet + t2 et .
Problem 16.13
y(t) = c1 sin 2t + c2 cos 2t − 1 + sin t.
Problem 16.14
t
Given that the function e5 satisfies the differential equation y 00 + 4y = et , write a
general solution of the differential equation y 00 + 4y = et .
Problem 16.15
Find the general solution to the differential equation
y (4) + 9y 00 = 24 + 108t2
given a particular solution yp (t) = cos 3t + sin 3t + t4 .
17 The Method of Variation of
Parameters
In this section, we discuss a method for finding a particular solution to a nonhomogeneous differential equation
y 00 + p(t)y 0 + q(t)y = g(t), a < t < b.
(17.1)
This method has no prior conditions to be satisfied by either p(t), q(t), or g(t).
To use this method, we first find the general solution to the homogeneous equation
y(t) = c1 y1 (t) + c2 y2 (t).
Then we replace the parameters c1 and c2 by two functions u1 (t) and u2 (t) to be
determined. From this the method got its name. Thus, obtaining
yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t).
Observe that if u1 and u2 are constant functions then the above y is just the
homogeneous solution to the differential equation.
In order to determine the two functions one has to impose two constraints. Finding
the derivative of yp we obtain
yp0 = (y10 u1 + y20 u2 ) + (y1 u01 + y2 u02 ).
Finding the second derivative to obtain
yp00 = y100 u1 + y10 u01 + y200 u2 + y20 u02 + (y1 u01 + y2 u02 )0 .
Since it is up to us to choose u1 and u2 we decide to do that in such a way to make
our computation simple. One way to achieving that is to impose the condition
y1 u01 + y2 u02 = 0.
123
(17.2)
124
17 THE METHOD OF VARIATION OF PARAMETERS
Under such a constraint yp0 and yp00 are simplified to
yp0 = y10 u1 + y20 u2
and
yp00 = y100 u1 + y10 u01 + y200 u2 + y20 u02 .
In particular, yp00 does not involve u001 and u002 .
Inserting yp , yp0 , and yp00 into equation (17.1) to obtain
[y100 u1 + y10 u01 + y200 u2 + y20 u02 ] + p(t)(y10 u1 + y20 u2 ) + q(t)(u1 y1 + u2 y2 ) = g(t).
Rearranging terms,
[y100 + p(t)y10 + q(t)y1 ]u1 + [y200 + p(t)y20 + q(t)y2 ]u2 + [u01 y10 + u02 y20 ] = g(t).
Since y1 and y2 are solutions to the homogeneous equation, the previous equation yields our second constraint
u01 y10 + u02 y20 = g(t).
(17.3)
Combining equation (17.2) and (17.3) we find the system of two equations in the
unknowns u01 and u02
y1 u01 + y2 u02 =0
u01 y10 + u02 y20 =g(t).
Since {y1 , y2 } is a fundamental set, the expression W (t) = y1 y20 − y10 y2 is nonzero
so that one can find unique u01 and u02 . Using the method of elimination, these
functions are given by
(t)g(t)
0
u01 (t) = − y2W
(t) and u2 (t) =
y1 (t)g(t)
W (t) .
Computing antiderivatives to obtain
R
R
(t)g(t)
u1 (t) = − y2W
dt
and
u
(t)
=
2
(t)
Example 17.1
Find the general solution of
y 00 − y 0 − 2y = 2e−t
using the method of variation of parameters.
y1 (t)g(t)
W (t) dt.
125
Solution.
The characteristic equation r2 − r − 2 = 0 has roots r1 = −1 and r2 = 2. Thus,
y1 (t) = e−t , y2 (t) = e2t and W (t) = 3et . Hence,
Z 2t
e · 2e−t
2
u1 (t) = −
dt = − t
t
3e
3
and
Z
u2 (t) =
e−t · 2e−t
2
dt = − e−3t .
t
3e
9
The particular solution is
2
2
yp (t) = − te−t − e−t .
3
9
The general solution is then given by
2
2
y(t) = c1 e−t + c2 e2t − te−t − e−t
3
9
Example 17.2
Find the general solution to (2t − 1)y 00 − 4ty 0 + 4y = (2t − 1)2 e−t if y1 (t) = t and
y2 (t) = e2t form a fundamental set of solutions to the equation.
Solution.
First we rewrite the equation in standard form
y 00 −
4t 0
4
y +
y = (2t − 1)e−t .
2t − 1
2t − 1
Since W (t) = (2t − 1)e2t we find
Z
u1 (t) = −
and
Z
u2 (t) =
e2t · (2t − 1)e−t
dt = e−t
(2t − 1)e2t
t · (2t − 1)e−t
1
1
dt = − te−3t − e−3t .
2t
(2t − 1)e
3
9
Thus,
1
1
2
1
yp (t) = te−t − te−t − e−t = te−t − e−t .
3
9
3
9
The general solution is
2
1
y(t) = c1 t + c2 e2t + te−t − e−t
3
9
126
17 THE METHOD OF VARIATION OF PARAMETERS
Example 17.3
Find the general solution to the differential equation y 00 + y 0 = ln t, t > 0.
Solution.
The characterisitc equation r2 + r = 0 has roots r1 = 0 and r2 = −1 so that
y1 (t) = 1, y2 (t) = e−t , and W (t) = −e−t . Hence,
Z −t
Z
e ln t
u1 (t) = −
dt
=
ln tdt = t ln t − t
−e−t
Z
Z
Z t
ln t
e
t
t
u2 (t) =
dt = − e ln tdt = −e ln t +
dt
−e−t
t
Thus,
yp (t) = t ln t − t − ln t + e
−t
Z
et
dt
t
and
y(t) = c1 + c2 e
−t
+ t ln t − t − ln t + e
−t
Z
et
dt
t
Example 17.4
Find the general solution of
y 00 + y =
1
.
2 + sin t
Solution.
Since the characteristic equation r2 + 1 = 0 has roots r = ±i, the general solution
of the corresponding homogeneous equation y 00 + y = 0 is given by
yh (t) = c1 cos t + c2 sin t
Since W (t) = 1 we find
Z
u1 (t) = −
Z
u2 (t) =
Z
sin t
2
dt = −t +
dt
2 + sin t
2 + sin t
cos t
dt = ln (2 + sin t)
2 + sin t
Hence, the particular solution is
Z
yp (t) = sin t ln (2 + sin t) + cos t(
2
dt − t)
2 + sin t
and the general solution is
y(t) = c1 cos t + c2 sin t + yp (t)
127
Practice Problems
Problem 17.1
Solve y 00 + y = sec t by variation of parameters.
Problem 17.2
Solve y 00 − y = et by variation of parameters.
Problem 17.3
Solve the following 2nd order equation using the variation of parameter method:
y 00 + 4y = t2 + 8 cos 2t.
Problem 17.4
Find a particular solution by the variation of parameters to the equation
y 00 + 2y 0 + y = e−t ln t.
Problem 17.5
Solve the following initial value problem by using variation of parameters:
y 00 + 2y 0 − 3y = tet , y(0) = −
1
59
, y 0 (0) = .
64
64
Problem 17.6 √
√
(a) Verify that {e t , e− t } is a fundamental set for the equation
4ty 00 + 2y 0 − y = 0
on the interval (0, ∞). You may assume that the given functions are solutions to
the equation.
(b) Use the method of variation of parameters to find one solution to the equation
√ √
4ty 00 + 2y 0 − y = 4 te t .
Problem 17.7
Use the method of variation of parameters to find the general solution to the
equation
y 00 + y = sin t.
128
17 THE METHOD OF VARIATION OF PARAMETERS
Problem 17.8
Consider the differential equation
t2 y 00 + 3ty 0 − 3y = 0, t > 0.
(a) Determine r so that y = tr is a solution.
(b) Use (a) to find a fundamental set of solutions.
(c) Use the method of variation of parameters for finding a particular solution to
t2 y 00 + 3ty 0 − 3y =
1
, t > 0.
t3
Problem 17.9
Use the method of variation of parameters to find the general solution to the
differential equation
y 00 + y = sin2 t.
18 The Laplace Transform:
Basic Definitions and Results
Laplace transform is yet another operational tool for solving constant coefficients
linear differential equations. The process of solution consists of three main steps:
• The given “hard” problem is transformed into a “simple” equation.
• This simple equation is solved by purely algebraic manipulations.
• The solution of the simple equation is transformed back to obtain the solution
of the given problem.
In this way the Laplace transformation reduces the problem of solving a differential
equation to an algebraic problem. The third step is made easier by tables, whose
role is similar to that of integral tables in integration.
The above procedure can be summarized by Figure 18.1.
Figure 18.1
In this section we introduce the concept of Laplace transform and discuss some of
its properties.
The Laplace transform is defined in the following way. Let f (t) be defined for
t ≥ 0. Then the Laplace transform of f, which is denoted by L[f (t)] or by F (s),
is defined by the following equation
Z T
Z ∞
−st
L[f (t)] = F (s) = lim
f (t)e dt =
f (t)e−st dt.
T →∞ 0
0
The integral which defines a Laplace transform is an improper integral. An improper integral may converge or diverge, depending on the integrand. When
the improper integral is convergent then we say that the function f (t) possesses
129
13018 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS
a Laplace transform. So what types of functions possess Laplace transforms, that
is, what type of functions guarantees a convergent improper integral?
Example 18.1
Find the Laplace transform, if it exists, of each of the following functions.
(a) f (t) = eat
2
(c) f (t) = t (d) f (t) = et .
(b) f (t) = 1
Solution.
(a) Using the definition of Laplace transform we see that
L[eat ] =
∞
Z
e−(s−a)t dt = lim
T →∞ 0
0
But
(
T
Z
T
Z
−(s−a)t
e
dt =
0
T
1−e−(s−a)T
s−a
e−(s−a)t dt.
if s = a
if s 6= a.
For the improper integral to converge we need s > a. In this case,
L[eat ] = F (s) =
1
, s > a.
s−a
(b) In a similar way to what was done in part (a), we find
∞
Z
L[1] =
−st
e
Z
dt = lim
T →∞ 0
0
T
1
e−st dt = , s > 0.
s
(c) We have
Z
L[t] =
∞
te
0
−st
te−st e−st
dt = −
− 2
s
s
∞
=
0
1
, s > 0.
s2
(d) Again using the definition of Laplace transform we find
Z ∞
2
t2
L[e ] =
et −st dt.
0
R
2 − st ≥ 0 so that et2 −st ≥ 1 and this implies that ∞ et2 −st dt ≥
If
s
≤
0
then
t
0
R∞
0 dt. Since the integral on the right is divergent, by the comparison theorem of
improper
R ∞ t(t−s) integrals
R ∞ the integral on the left is also divergent. Now, if s > 0 then
dt ≥ s dt. By the same reasoning the integral on the left is divergent.
0 e
2
This shows that the function f (t) = et does not possess a Laplace transform
131
The above example raises the question of what class or classes of functions possess
a Laplace transform.
Looking closely at Example 18.1(a), we notice that for s > a
R∞
the integral 0 e−(s−a)t dt is convergent and a critical component for this convergence is the type of the function f (t). To be more specific, if f (t) is a continuous
function such that
|f (t)| ≤ M eat ,
t≥C
(18.1)
where M ≥ 0 and a and C are constants, then this condition yields
Z
∞
f (t)e
0
−st
Z
dt ≤
C
−st
f (t)e
0
Z
∞
dt + M
e−(s−a)t dt.
C
Since f (t) is continuous in 0 ≤ t ≤ C, by letting A = max{|f (t)| : 0 ≤ t ≤ C} we
have
Z C
Z C
1 e−sC
f (t)e−st dt ≤ A
e−st dt = A
−
< ∞.
s
s
0
0
R∞
Also, by Example 18.1(a), the integral C e−(s−a)t dt is convergent for s > a.
By the comparison theorem of improper integrals the integral on the left is also
convergent. That is, f (t) possesses a Laplace transform.
We call a function that satisfies condition (18.1) a function with an exponential
order a. Graphically, this means that the graph of f (t) is contained in the region
bounded by the graphs of y = M eat and y = −M eat for t ≥ C. Note also that
this type of functions controls the negative exponential in the transform integral
so that to keep the integral from blowing up. If C = 0 then we say that the
function is exponentially bounded. Note that f has exponential order a if
limt→∞ e−at f (t) = 0.
Example 18.2
(a) Show that f (t) = tn where n is a positive integer, has an exponential order.
(b) Show that any bounded function f (t) for t ≥ 0 is exponentially bounded.
2
(c) Show that f (t) = et is not of exponential order.
Solution.
(a) Let a > 0. Using L’Hôpital’s rule n times we can see that limt→∞ e−at tn = 0.
Thus, there is a positive constant C such that e−at tn ≤ 1 for all t ≥ C. That is,
tn ≤ eat for all t ≥ C.
(b) Since f (t) is bounded for t ≥ 0, there is a positive constant M such that
|f (t)| ≤ M for all t ≥ 0. But this is the same as (18.1) with a = 0 and C = 0.
Thus, f (t) is exponentially bounded.
(c) Suppose not. That is, let M and C be non-negative constants such that
13218 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS
2
2
et ≤ M eat for some constant a and for all t ≥ C. Hence, et −at ≤ M for all t ≥ C.
Letting t → ∞ we find M ≥ ∞ which is impossible since M < ∞. It follows that
2
f (t) = et is not of exponential order
Another question that comes to mind is whether it is possible to relax the condition
of continuity on the function f (t). Let’s look at the following situation.
Example 18.3
Show that the square wave function whose graph is given in Figure 18.2 possesses
a Laplace transform.
Figure 18.2
Note that the function is periodic of period 2.
Solution.
R∞
R∞
Since f (t)e−st ≤ e−st , 0 f (t)e−st dt ≤ 0 e−st dt. But the integral on the right is
convergent for s > 0 so that the integral on the left is convergent as well. That is,
L[f (t)] exists for s > 0
The function of the above example belongs to a class of functions that we define
next.
A function is called piecewise continuous on an interval if it consists of a finite
number of continuous pieces with possibly either removable or jump discontinuities (but no inifinite discontinuities). Below is a sketch of a piecewise continuous
function.
Figure 18.3
An example of a function that is not piecewise continuous is the function f (t) =
on the interval [0, ∞) since at t = 1 the continuity is infinite.
1
t−1
133
Example 18.4
Show that the following function is piecewise continuous and exponentially bounded.
f (t) =
et sin (t−1)
1
2
if t ≥ 1
if 0 ≤ t < 1.
Solution.
Since et sin (t−1) = (esin (t−1) )t ≤ (e1 )t = et for all t ≥ 0, f (t) is piecewise continuous
and exponentially bounded
The following theorem guarantees the existence of the Laplace transform for all
functions that are piecewise continuous and have exponential order.
Theorem 18.1 (Existence)
Suppose that f (t) is piecewise continuous on t ≥ 0 and has an exponential order
with |f (t)| ≤ M eat for t ≥ C. Then the Laplace transform
Z
F (s) =
∞
f (t)e−st dt
0
exists as long as s > a. Note that the two conditions above are sufficient, but not
necessary, for F (s) to exist.
Example 18.5
Consider the floor function f (t) = btc, where for any integer n we have btc = n for
all n ≤ t < n + 1.
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on
0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Solution.
(a) The floor function is a piecewise continuous function on 0 ≤ t < ∞.
(b) Since btc ≤ t < et for 0 ≤ t < ∞ we find M = 1 and a = 1
Example 18.6
Consider the function f (t) = t2 e−t .
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on
0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
13418 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS
Solution.
(a) Since t2 and e−t are continuous everywhere, f (t) = t2 e−t is continuous on
0 ≤ t < ∞.
(b) By L’Hôpital’s rule one has
t2
= 0.
t→∞ et
lim
Since f (0) = 0, f (t) is bounded. Since f 0 (t) = (2t − t2 )e−t , f (t) has a maximum
when t = 2. The value of this maximum is f (2) = 4e−2 . Hence, M = 4e−2 and
a=0
In what follows, we will denote the class of all piecewise continuous functions
with exponential order by PE. The next theorem shows that any linear combination of functions in PE is also in PE. The same is true for the product of two
functions in PE.
Theorem 18.2
Suppose that f (t) and g(t) are two elements of PE with
|f (t)| ≤ M1 ea1 t ,
t ≥ C1
and
|g(t)| ≤ M2 ea1 t ,
t ≥ C2 .
(i) For any constants α and β the function αf (t) + βg(t) is also a member of PE.
Moreover
L[αf (t) + βg(t)] = αL[f (t)] + βL[g(t)].
(ii) The function h(t) = f (t)g(t) is an element of PE.
Example 18.7
Use the linearity property of Laplace transform to find L[5e−7t + t + 2e2t ]. Find
the domain of F (s).
Solution.
We have L[e−7t ] =
1
s+7 ,
s > −7, L[t] =
1
,
s2
s > 0, and L[e2t ] =
L[5e−7t + t + 2e2t ] = 5L[e−7t ] + L[t] + 2L[e2t ] =
1
s−2 ,
s > 2. Hence,
5
1
2
+ 2+
, s>2
s+7 s
s−2
We next discuss the problem of how to determine the function f (t) if F (s) is given.
That is, how do we invert the transform. The following result on uniqueness provides a possible answer. This result establishes a one-to-one correspondence between the set PE and its Laplace transforms. Alternatively, the following theorem
asserts that the Laplace transform of a member in PE is unique.
135
Theorem 18.3
Let f (t) and g(t) be two elements in PE with Laplace transforms F (s) and G(s)
such that F (s) = G(s) for some s > a. Then f (t) = g(t) for all t ≥ 0 where both
functions are continuous.
With the above theorem, we can now officially define the inverse Laplace transform
as follows: For a piecewise continuous function f with exponential order whose
Laplace transform is F, we call f the inverse Laplace transform of F and write
f = L−1 [F (s)]. Symbolically
f (t) = L−1 [F (s)] ⇐⇒ F (s) = L[f (t)].
Example18.8
1
Find L−1 s−1
, s > 1.
Solution.
1
, s > a. In particular, for a = 1
From Example 18.1(a), we have that L[eat ] = s−a
1
1
we find that L[et ] = s−1
, s > 1. Hence, L−1 s−1
= et , t ≥ 0
The above theorem states that if f (t) is continuous and has a Laplace transform
F (s), then there is no other function that has the same Laplace transform. To
find L−1 [F (s)], we can inspect tables of Laplace transforms of known functions to
find a particular f (t) that yields the given F (s).
When the function f (t) is not continuous, the uniqueness of the inverse Laplace
transform is not assured. The following example addresses the uniqueness issue.
Example 18.9
Consider the two functions f (t) = h(t)h(3 − t) and g(t) = h(t) − h(t − 3) where
1
if t ≥ 0
h(t) =
0 otherwise.
(a) Are the two functions identical?
(b) Show that L[f (t)] = L[g(t).
Solution.
(a) We have
1, 0 ≤ t ≤ 3
0,
t>3
1, 0 ≤ t < 3
0,
t ≥ 3.
f (t) =
and
g(t) =
13618 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS
Since f (3) = 1 and g(3) = 0, f and g are not identical.
(b) We have
Z
L[f (t)] = L[g(t)] =
0
3
e−st dt =
1 − e−3s
, s > 0.
s
Thus, both functions f (t) and g(t) have the same Laplace transform even though
they are not identical. However, they are equal on the interval(s) where they are
both continuous
The inverse Laplace transform possesses a linear property as indicated in the following result.
Theorem 18.4
Given two Laplace transforms F (s) and G(s) then
L−1 [aF (s) + bG(s)] = aL−1 [F (s)] + bL−1 [G(s)]
for any constants a and b.
Example18.10
2
Find L−1 s+2
+
2
s−2
.
Solution.
We have
2
2
1
1
L−1
+
= 2L−1
+ 2L−1
= 2(e−2t + e2t ), t ≥ 0
s+2 s−2
s+2
s−2
137
Practice Problems
Problem 18.1
R∞
Determine whether the integral 0
give its value.
1
dt
1+t2
converges. If the integral converges,
Problem 18.2
R∞
Determine whether the integral 0
give its value.
t
dt
1+t2
converges. If the integral converges,
Problem 18.3
R∞
Determine whether the integral 0 e−t cos (e−t )dt converges. If the integral converges, give its value.
Problem 18.4
Using the definition, find L[e3t ], if it exists. If the Laplace transform exists then
find the domain of F (s).
Problem 18.5
Using the definition, find L[t − 5], if it exists. If the Laplace transform exists then
find the domain of F (s).
Problem 18.6
2
Using the definition, find L[e(t−1) ], if it exists. If the Laplace transform exists
then find the domain of F (s).
Problem 18.7
Using the definition, find L[(t − 2)2 ], if it exists. If the Laplace transform exists
then find the domain of F (s).
Problem 18.8
Using the definition, find L[f (t)], if it exists. If the Laplace transform exists then
find the domain of F (s).
0,
0≤t<1
f (t) =
t − 1,
t ≥ 1.
Problem 18.9
Using the definition, find L[f (t)], if it exists. If the Laplace transform exists then
find the domain of F (s).

0≤t<1
 0,
t − 1, 1 ≤ t < 2
f (t) =

0,
t ≥ 2.
13818 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS
Problem 18.10
Let n be a positive integer. Using integration by parts establish the reduction
formula
Z
Z
tn e−st n
tn e−st dt = −
+
tn−1 e−st dt, s > 0.
s
s
Problem 18.11
For s > 0 and n a positive integer evaluate the limits
(a) limt→0 tn e−st
(b) limt→∞ tn e−st .
Problem 18.12
(a) Use the previous two problems to derive the reduction formula for the Laplace
transform of f (t) = tn ,
n
L[tn ] = L[tn−1 ], s > 0.
s
(b) Calculate L[tk ], for k = 1, 2, 3, 4, 5.
(c) Formulate a conjecture as to the Laplace transform of f (t), tn with n a positive
integer.
From a table of integrals,
Z
α sin βu − β cos βu
eαu sin βudu =eαu
α2 + β 2
Z
α cos βu + β sin βu
eαu cos βudu =eαu
α2 + β 2
Problem 18.13
Use the above integrals to find the Laplace transform of f (t) = cos ωt, if it exists.
If the Laplace transform exists, give the domain of F (s).
Problem 18.14
Use the above integrals to find the Laplace transform of f (t) = sin ωt, if it exists.
If the Laplace transform exists, give the domain of F (s).
Problem 18.15
Use the above integrals to find the Laplace transform of f (t) = cos ω(t − 2), if it
exists. If the Laplace transform exists, give the domain of F (s).
Problem 18.16
Use the above integrals to find the Laplace transform of f (t) = e3t sin t, if it exists.
If the Laplace transform exists, give the domain of F (s).
139
Problem 18.17
Consider the function f (t) = tan t.
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on
0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Problem 18.18
t2
Consider the function f (t) = e2te +1 .
(a) Is f (t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on
0 ≤ t < ∞, or neither?
(b) Are there fixed numbers a and M such that |f (t)| ≤ M eat for 0 ≤ t < ∞?
Problem18.19
3
Find L−1 s−2
.
Problem18.20
Find L−1 − s22 +
1
s+1
.
14018 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS
19 Further Studies of Laplace
Transform
Properties of the Laplace transform enable us to find Laplace transforms without
having to compute them directly from the definition. In this section, we establish
properties of Laplace transform that will be useful for solving ODEs.
Laplace Transform of the Heaviside Step Function
The Heaviside step function is a piecewise continuous function defined by
h(t) =
1, t ≥ 0
0, t < 0.
Figure 19.1 displays the graph of h(t).
Figure 19.1
Taking the Laplace transform of h(t) we find
Z
L[h(t)] =
∞
h(t)e
0
−st
Z
dt =
∞
−st
e
0
141
−st ∞
e
1
dt = −
= , s > 0.
s 0
s
142
19 FURTHER STUDIES OF LAPLACE TRANSFORM
A Heaviside function at α ≥ 0 is the shifted function h(t − α) (α units to the
right). For this function, the Laplace transform is
−st ∞
Z ∞
Z ∞
e
e−sα
−st
−st
L[h(t − α)] =
h(t − α)e dt =
e dt = −
=
, s > 0.
s α
s
0
α
Laplace Tranform of eat
The Laplace transform for the function f (t) = eat is
#∞
"
Z ∞
1
e−(s−a)t
−(s−a)t
at
=
, s > a.
e
dt = −
L[e ] =
s−a
s−a
0
0
Laplace Tranforms of sin at and cos at
Using integration by parts twice we find
Z ∞
L[sin at] =
e−st sin atdt
0
−st
∞
Z
a2 ∞ −st
e sin at ae−st cos at
−
= −
−
e sin atdt
s
s2
s2 0
0
=−
s2 + a2
s2
L[sin at] =
L[sin at] =
a
a2
−
L[sin at]
s2
s2
a
s2
s2
a
, s > 0.
+ a2
A similar argument shows that
L[cos at] =
s
, s > 0.
s2 + a2
Laplace Transforms of cosh at and sinh at
Using the linear property of L we can write
1
L[eat ] + L[e−at ]
2
1
1
1
=
+
, s > |a|
2 s−a s+a
s
, s > |a|.
= 2
s − a2
L[cosh at] =
A similar argument shows that
L[sinh at] =
a
, s > |a|.
s2 − a2
143
Laplace Transform of a Polynomial
Let n be a positive integer. Using integration by parts we can write
Z
∞
n −st
t e
0
tn e−st
dt = −
s
∞
0
n
+
s
Z
∞
tn−1 e−st dt.
0
By repeated use of L’Hôpital’s rule we find limt→∞ tn e−st = limt→∞
s > 0. Thus,
n
L[tn ] = L[tn−1 ], s > 0.
s
n!
sn est
= 0 for
Using induction on n = 0, 1, 2, · · · one can easily establish that
L[tn ] =
n!
, s > 0.
sn+1
Using the above result together with the linearity property of L one can find the
Laplace transform of any polynomial.
The next two results are referred to as the first and second shift theorems. As
with the linearity property, the shift theorems increase the number of functions
for which we can easily find Laplace transforms.
Theorem 19.1 (First Shifting Theorem)
If f (t) is a piecewise continuous function for t ≥ 0 and has exponential order with
|f (t)| ≤ M eat , t ≥ C, then for any real number α we have
L[eαt f (t)] = F (s − α), s > a + α
where L[f (t)] = F (s).
Theorem 19.2 (Second Shifting Theorem)
If f (t) is a piecewise continuous function for t ≥ 0 and has exponential order with
|f (t)| ≤ M eat , t ≥ C, then for any real number α ≥ 0 we have
L[f (t − α)h(t − α)] = e−αs F (s), s > a
where L[f (t)] = F (s) and h(t) is the Heaviside step function.
Example 19.1
Find
(a) L[e2t t2 ]
(b) L[e3t cos 2t]
−2s
(c) L−1 [ e s2 ].
144
19 FURTHER STUDIES OF LAPLACE TRANSFORM
Solution.
(a) By Theorem 19.1, we have L[e2t t2 ] = F (s − 2) where L[t2 ] = s2!3 = F (s), s > 0.
2
Thus, L[e2t t2 ] = (s−2)
3 , s > 2.
(b) As in part (a), we have L[e3t cos 2t] = F (s − 3) where L[cos 2t] = F (s). But
L[cos 2t] = s2s+4 , s > 0. Thus,
L[e3t cos 2t] =
1
,
s2
(c) Since L[t] =
s−3
, s > 3.
(s − 3)2 + 4
by Theorem 19.2, we have
e−2s
= L[(t − 2)h(t − 2)].
s2
Therefore,
−1
L
e−2s
0,
0≤t<2
= (t − 2)h(t − 2) =
t − 2, t ≥ 2
s2
The following result relates the Laplace transform of derivatives and integrals to
the Laplace transform of the function itself.
Theorem 19.3
Suppose that f (t) is continuous for t ≥ 0 and f 0 (t) is piecewise continuous and
exponentially bounded. Then
(a) f (t) is exponentially bounded.
(b) L[f 0 (t)] = sL[f (t)] − f (0) = sF (s) − f (0), s > max{a, 0} + 1.
0
2
(c) L[fh 00 (t)] = s2 L[f
i (t)]−sf (0)−f (0) = s F (s)−sf (0)−f (0), s > max{a, 0}+1.
Rt
(d) L 0 f (u)du = L[fs(t)] = F (s)
s , s > max{a, 0} + 1.
The results in (b) and (c) of the previous theorem can be extended to higher order
derivatives.
Theorem 19.4
Let f (t), f 0 (t), · · · , f (n−1) (t) be continuous and f (n) (t) be piecewise continuous and
exponentially bounded. Then
L[f (n) (t)] = sn L[f (t)] − sn−1 f (0) − sn−2 f 0 (0) − · · · − f (n−1) (0), s > max{a, 0} + 1.
We next illustrate the use of the previous theorem in solving initial value problems.
145
Example 19.2
Use Laplace transform technique to solve the initial value problem
y 00 − 4y = e3t , y(0) = 0, y 0 (0) = 0.
Solution.
Taking the Laplace transform of the DE and using linearity we find
L[y 00 ] − 4L[y] = L[e3t ].
But L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 L[y]. Letting L[y] = Y (s) we obtain
s2 Y (s) − 4Y (s) =
1
.
s−3
Solving for Y (s) we find
Y (s) =
1
.
(s − 3)(s − 2)(s + 2)
Using partial fraction decomposition
1
A
B
C
=
+
+
(s − 3)(s − 2)(s + 2)
s−3 s+2 s−2
1
we find A = 15 , B = 20
, and C = − 14 . Thus,
1
1 −1
1
1 −1
1
1 −1
1
−1
y(t) =L
= L
+ L
− L
(s − 3)(s − 2)(s + 2)
5
s−3
20
s+2
4
s−2
1
1
1
= e3t + e−2t − e2t , t ≥ 0
5
20
4
Example 19.3
Determine the constants α, β, y0 , and y00 so that Y (s) =
transform of the solution to the initial value problem
s
(s+1)2
y 00 + αy 0 + βy = 0, y(0) = y0 , y 0 (0) = y00 .
Solution.
Taking the Laplace transform of both sides we find
s2 Y (s) − sy0 − y00 + αsY (s) − αy0 + βY (s) = 0.
Solving for Y (s) we find
Y (s) =
sy0 + (y00 + αy0 )
s
= 2
.
s2 + αs + β
s + 2s + 1
By identification we find α = 2, β = 1, y0 = 1, and y00 = −2
is the Laplace
146
19 FURTHER STUDIES OF LAPLACE TRANSFORM
Example 19.4
Solve the initial value problem
y 00 − 4y 0 + 9y = t, y(0) = 0, y 0 (0) = 1.
Solution.
We apply Theorem 19.4 that gives the Laplace transform of a derivative. By the
linearity property of the Laplace transform we can write
L[y 00 ] − 4L[y 0 ] + 9L[y] = L[t].
Now since
L[y 00 ] =s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − 1
L[y 0 ] =sY (s) − y(0) = sY (s)
1
L[t] = 2
s
where L[y] = Y (s), we obtain
s2 Y (s) − 1 − 4sY (s) + 9Y (s) =
1
.
s2
Rearranging gives
(s2 − 4s + 9)Y (s) =
s2 + 1
.
s2
Thus,
Y (s) =
and
y(t) = L−1
s2 + 1
s2 (s2 − 4s + 9)
s2 + 1
s2 (s2 − 4s + 9)
In the next section we will discuss a method for finding the inverse Laplace transform of the above expression.
147
f(t)
F(s)
h(t) =
1, t ≥ 0
0, t < 0
1
s,
s>0
tn , n = 1, 2, · · ·
n!
,
sn+1
s>0
eαt
1
s−α ,
s>α
sin (ωt)
ω
,
s2 +ω 2
s>0
cos (ωt)
s
,
s2 +ω 2
s>0
sinh (ωt)
ω
,
s2 −ω 2
s > |ω|
cosh (ωt)
s
,
s2 −ω 2
s > |ω|
eαt f (t), with |f (t)| ≤ M eat
F (s − α), s > α + a
eαt h(t)
1
s−α ,
eαt tn , n = 1, 2, · · ·
n!
,
(s−α)n+1
eαt sin (ωt)
ω
,
(s−α)2 +ω 2
s>α
eαt cos (ωt)
s−α
,
(s−α)2 +ω 2
s>α
f (t − α)h(t − α), α ≥ 0
with |f (t)| ≤ M eat
e−αs F (s), s > a
s>α
s>α
148
19 FURTHER STUDIES OF LAPLACE TRANSFORM
f(t)
h(t − α), α ≥ 0
F(s) (continued)
e−αs
s , s>0
tf (t)
-F 0 (s)
t
2ω
s
,
(s2 +ω 2 )2
s>0
1
,
(s2 +ω 2 )2
s>0
sin ωt
1
[sin ωt
2ω 3
− ωt cos ωt]
f 0 (t), with f (t) continuous
and |f 0 (t)| ≤ M eat
sF (s) − f (0)
s > max{a, 0} + 1
f 00 (t), with f 0 (t) continuous
and |f 00 (t)| ≤ M eat
s2 F (s) − sf (0) − f 0 (0)
s > max{a, 0} + 1
f (n) (t), with f (n−1) (t) continuous
and |f (n) (t)| ≤ M eat
sn F (s) − sn−1 f (0) − · · ·
-sf (n−2) (0) − f (n−1) (0)
s > max{a, 0} + 1
Rt
F (s)
s ,
0
f (u)du, with |f (t)| ≤ M eat
Table L
s > max{a, 0} + 1
149
Practice Problems
Problem 19.1
Use Table L to find L[2et + 5].
Problem 19.2
Use Table L to find L[e3t−3 h(t − 1)].
Problem 19.3
Use Table L to find L[sin2 ωt].
Problem 19.4
Use Table L to find L[sin 3t cos 3t].
Problem 19.5
Use Table L to find L[e2t cos 3t].
Problem 19.6
Use Table L to find L[e4t (t2 + 3t + 5)].
Problem 19.7
Use Table L to find L−1 [ s210
+
+25
4
s−3 ].
Problem 19.8
5
Use Table L to find L−1 [ (s−3)
4 ].
Problem 19.9
−2s
Use Table L to find L−1 [ es−9 ].
Problem 19.10
−3s
Use Table L to find L−1 [ e s2(2s+7)
].
+16
Problem 19.11
Graph the function f (t) = h(t − 1) + h(t − 3) for t ≥ 0, where h(t) is the Heaviside
step function, and use Table L to find L[f (t)].
Problem 19.12
Graph the function f (t) = t[h(t−1)−h(t−3)] for t ≥ 0, where h(t) is the Heaviside
step function, and use Table L to find L[f (t)].
Problem 19.13
Graph the function f (t) = 3[h(t−1)−h(t−4)] for t ≥ 0, where h(t) is the Heaviside
step function, and use Table L to find L[f (t)].
150
19 FURTHER STUDIES OF LAPLACE TRANSFORM
Problem 19.14
Graph the function f (t) = |2 − t|[h(t − 1) − h(t − 3)] for t ≥ 0, where h(t) is the
Heaviside step function, and use Table L to find L[f (t)].
Problem 19.15
Graph the function f (t) = h(2 − t) for t ≥ 0, where h(t) is the Heaviside step
function, and use Table L to find L[f (t)].
Problem 19.16
Graph the function f (t) = h(t − 1) + h(4 − t) for t ≥ 0, where h(t) is the Heaviside
step function, and use Table L to find L[f (t)].
Problem 19.17
The graph of f (t) is given below. Represent f (t) as a combination of Heaviside
step functions, and use Table L to calculate the Laplace transform of f (t).
Problem 19.18
The graph of f (t) is given below. Represent f (t) as a combination of Heaviside
step functions, and use Table L to calculate the Laplace transform of f (t).
Problem 19.19
Use Laplace transform technique to solve the initial value problem
y 0 + 4y = g(t), y(0) = 2
where

 0, 0 ≤ t < 1
12, 1 ≤ t < 3
g(t) =

0,
t ≥ 3.
20 The Laplace Transform and
the Method of Partial Fractions
In the last example of the previous section we encountered the equation
s2 + 1
−1
y(t) = L
.
s2 (s2 − 4s + 9)
We would like to find an explicit expression for y(t). This can be done using the
method of partial fractions
which is the topic of this section. According to this
N (s)
−1
method, finding L
, where N (s) and D(s) are polynomials, require deD(s)
composing the rational function into a sum of simpler expressions whose inverse
Laplace transform can be recognized from a table of Laplace transform pairs.
The method of partial fractions decomposition consists of writing a rational function, i.e., function of the form
R(s) =
N (s)
D(s)
where N (s) and D(s) are polynomials, as a sum of simpler fractions called partial
fractions. This can be done in the following way:
Step 1. Use long division to find two polynomials r(s) and q(s) such that
N (s)
r(s)
= q(s) +
.
D(s)
D(s)
Note that if the degree of N (s) is smaller than that of D(s) then q(s) = 0 and
r(s) = N (s).
Step 2. Write D(s) as a product of factors of the form (as + b)n or (as2 + bs + c)n
where as2 + bs + c is irreducible, i.e. as2 + bs + c = 0 has no real zeros.
151
15220 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS
Step 3. Decompose
r(s)
D(s)
into a sum of partial fractions in the following way:
(1) For each factor of the form (as + b)k write
A1
A2
Ak
+
+ ··· +
,
2
as + b (as + b)
(as + b)k
where the numbers A1 , A2 , · · · , Ak are to be determined.
(2) For each factor of the form (as2 + bs + c)k write
B 1 s + C1
B k s + Ck
B2 s + C2
+ ··· +
,
+
2
2
2
as + bs + c (as + bs + c)
(as2 + bs + c)k
where the numbers B1 , B2 , · · · , Bk and C1 , C2 , · · · , Ck are to be determined.
Step 4. Multiply both sides of
sion of the form
r(s)
D(s)
by D(s) and simplify. This leads to an expres-
r(s) = a polynomial whose coefficients are combinations of Ai , Bi , and Ci .
Finally, we find the constants, Ai , Bi , and Ci by equating the coefficients of like
powers of x on both sides of the last equation.
Example 20.1
Decompose into partial fractions R(s) =
s3 +s2 +2
.
s2 −1
Solution.
3
2 +2
= s + 1 + ss+3
Step 1. s s+s
2 −1
2 −1 .
2
Step 2. s − 1 = (s − 1)(s + 1).
B
s+3
A
+ s−1
.
Step 3. (s+1)(s−1)
= s+1
Step 4. Multiply both sides of the last equation by (s − 1)(s + 1) to obtain
s + 3 = A(s − 1) + B(s + 1).
Expand the right hand side, collect terms with the same power of s, and identify
coefficients of the polynomials obtained on both sides:
s + 3 = (A + B)s + (B − A).
Hence, A + B = 1 and B − A = 3. Adding these two equations gives B = 2. Thus,
A = −1 and so
s3 + s2 + 2
1
2
=s+1−
+
2
s −1
s+1 s−1
Now, after decomposing the rational function into a sum of partial fractions all we
A
need to do is to find the Laplace transform of expressions of the form (s−α)
n or
Bs+C
.
(as2 +bs+c)n
153
Exampleh20.2 i
1
Find L−1 s(s−3)
.
Solution.
We write
1
A
B
= +
.
s(s − 3)
s
s−3
Multiply both sides by s(s − 3) and simplify to obtain
1 = A(s − 3) + Bs
or
1 = (A + B)s − 3A.
Now equating the coefficients of like powers of s to obtain −3A = 1 and A+B = 0.
Solving for A and B we find A = − 13 and B = 13 . Thus,
1
1 −1 1
1 −1
1
−1
L
=− L
+ L
s(s − 3)
3
s
3
s−3
1
1 3t
= − h(t) + e , t ≥ 0
3
3
where h(t) is the Heaviside unit step function
Exampleh20.3 i
Find L−1 s3s+6
2 +3s .
Solution.
We factor the denominator and split the integrand into partial fractions:
3s + 6
A
B
= +
.
s(s + 3)
s
s+3
Multiplying both sides by s(s + 3) to obtain
3s + 6 = A(s + 3) + Bs
= (A + B)s + 3A
Equating the coefficients of like powers of x to obtain 3A = 6 and A + B = 3.
Thus, A = 2 and B = 1. Finally,
1
1
3s + 6
L−1 2
= 2L−1
+ L−1
= 2h(t) + e−3t , t ≥ 0.
s + 3s
s
s+3
15420 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS
Exampleh20.4 i
s2 +1
Find L−1 s(s+1)
2 .
Solution.
We factor the denominator and split the rational function into partial fractions:
s2 + 1
A
B
C
= +
.
+
2
s(s + 1)
s
s + 1 (s + 1)2
Multiplying both sides by s(s + 1)2 and simplifying to obtain
s2 + 1 =A(s + 1)2 + Bs(s + 1) + Cs
=(A + B)s2 + (2A + B + C)s + A.
Equating coefficients of like powers of s we find A = 1, 2A + B + C = 0 and
A + B = 1. Thus, B = 0 and C = −2. Now finding the inverse Laplace transform
to obtain
2
s +1
1
−1
−1 1
−1
L
=L
− 2L
= h(t) − 2te−t , t ≥ 0.
s(s + 1)2
s
(s + 1)2
Example 20.5
Use Laplace transform to solve the initial value problem
y 00 + 3y 0 + 2y = e−t , y(0) = y 0 (0) = 0.
Solution.
By the linearity property of the Laplace transform we can write
L[y 00 ] + 3L[y 0 ] + 2L[y] = L[e−t ].
Now since
L[y 00 ] =s2 L[y] − sy(0) − y 0 (0) = s2 Y (s)
L[y 0 ] =sY (s) − y(0) = sY (s)
1
L[e−t ] =
s+1
where L[y] = Y (s), we obtain
s2 Y (s) + 3sY (s) + 2Y (s) =
1
.
s+1
155
Rearranging gives
(s2 + 3s + 2)Y (s) =
1
.
s+1
Thus,
Y (s) =
and
−1
y(t) = L
(s +
1
.
+ 3s + 2)
1)(s2
1
.
(s + 1)(s2 + 3s + 2)
Using the method of partial fractions we can write
(s +
1
1
1
1
=
−
+
.
+ 3s + 2)
s + 2 s + 1 (s + 1)2
1)(s2
Thus,
−1
y(t) = L
1
1
1
−1
−1
−L
+L
= e−2t − e−t + te−t , t ≥ 0
s+2
s+1
(s + 1)2
15620 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS
Practice Problems
In Problems 20.1 - 20.4, give the form of the partial fraction expansion for F (s).
You need not evaluate the constants in the expansion. However, if the denominator has an irreducible quadratic expression then use the completing the square
process to write it as the sum/difference of two squares.
Problem 20.1
F (s) =
s3 + 3s + 1
.
(s − 1)3 (s − 2)2
Problem 20.2
F (s) =
s2 + 5s − 3
.
+ 16)(s − 2)
(s2
Problem 20.3
F (s) =
s3 − 1
.
(s2 + 1)2 (s + 4)2
Problem 20.4
F (s) =
s4 + 5s2 + 2s − 9
.
(s2 + 8s + 17)(s − 2)2
Problemh20.5 i
1
Find L−1 (s+1)
3 .
Problemh20.6 i
Find L−1 s22s−3
.
−3s+2
Problemh20.7 i
2
Find L−1 4ss3+s+1
.
+s
Problemh20.8
i
2 +6s+8
Find L−1 s4s+8s
2 +16 .
Problem 20.9
Use Laplace transform to solve the initial value problem
y 0 + 2y = 26 sin 3t, y(0) = 3.
157
Problem 20.10
Use Laplace transform to solve the initial value problem
y 0 + 2y = 4t, y(0) = 3.
Problem 20.11
Use Laplace transform to solve the initial value problem
y 00 + 3y 0 + 2y = 6e−t , y(0) = 1, y 0 (0) = 2.
Problem 20.12
Use Laplace transform to solve the initial value problem
y 00 + 4y = cos 2t, y(0) = 1, y 0 (0) = 1.
Problem 20.13
Use Laplace transform to solve the initial value problem
y 00 − 2y 0 + y = e2t , y(0) = 0, y 0 (0) = 0.
Problem 20.14
Use Laplace transform to solve the initial value problem
y 00 + 9y = g(t), y(0) = 1, y 0 (0) = 3
where
g(t) =
6, 0 ≤ t < π
0, π ≤ t < ∞
Problem 20.15
Determine the constants α, β, y0 , and y00 so that Y (s) =
transform of the solution to the initial value problem
2s−1
s2 +s+2
y 00 + αy 0 + βy = 0, y(0) = y0 , y 0 (0) = y00 .
is the Laplace
15820 THE LAPLACE TRANSFORM AND THE METHOD OF PARTIAL FRACTIONS
21 Laplace Transforms of
Periodic Functions
In many applications, the nonhomogeneous term in a linear differential equation is
a periodic function. In this section, we derive a formula for the Laplace transform
of such periodic functions.
Recall that a function f (t) is said to be T −periodic if T is the smallest positive
interger such that f (t + T ) = f (t) whenever t and t + T are in the domain of f (t).
For example, the sine and cosine functions are 2π−periodic whereas the tangent
and cotangent functions are π−periodic.
If f (t) is T −periodic for t ≥ 0 then we define the function
f (t), 0 ≤ t ≤ T
fT (t) =
0,
t > T.
The Laplace transform of this function is then
Z
Z ∞
−st
fT (t)e dt =
L[fT (t)] =
T
f (t)e−st dt.
0
0
The Laplace transform of a T −periodic function is given next.
Theorem 21.1
If f (t) is a T −periodic, piecewise continuous function for t ≥ 0 then
L[f (t)] =
L[fT (t)]
, s > 0.
1 − e−sT
Example 21.1
Determine the Laplace transform of the function

 1, 0 ≤ t ≤ T2
f (t + T ) = f (t), t ≥ 0.
f (t) =

0, T2 < t < T.
159
160
21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS
Solution.
The graph of f (t) is shown in Figure 21.1.
Figure 21.1
We have
Z
L[fT (t)] =
T
2
e−st dt
0
1 − e−
=
s
sT
2
.
By Theorem 21.1, we have
sT
L[f (t)] =
1−e− 2
s
1 − e−sT
=
1
s(1 + e−
sT
2
, s>0
)
Example 21.2
Find the Laplace transform of the sawtooth curve shown in Figure 21.2
Figure 21.2
161
Solution.
The given function is periodic of period b. For the first period the function is
defined by
a
fb (t) = t[h(t) − h(t − b)].
b
So we have
Z b
a
e−st tdt
L[fb (t)] =
b
0
b
t −st e−st
a
− 2
= − e
b
s
s 0
−bs
a 1
bse
+ e−bs
=
−
.
b s2
s2
Hence,
L[fb (t)]
a 1 − e−bs − bse−bs
L[f (t)] =
=
b
1 − e−bs
s2 (1 − e−bs )
Exampleh21.3
i
e−s
Find L−1 s12 − s(1−e
−s ) .
Solution.
Note first that
1
e−s
1 − e−s − se−s
−
=
.
s2 s(1 − e−s )
s2 (1 − e−s )
According to the previous example with a = 1 and b = 1 we find that L−1
is the sawtooth function shown in Figure 21.2
h
1
s2
−
e−s
s(1−e−s )
Linear Time Invariant Systems and the Transfer Function
The Laplace transform is a powerful technique for analyzing linear time-invariant
systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems, to name just a few. A mathematical model described by a linear
differential equation with constant coefficients of the form
an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = bm u(m) + bm−1 u(m−1) + · · · + b1 u0 + b0 u
is called a linear time invariant system. The function y(t) denotes the system
output and the function u(t) denotes the system input. The system is called timeinvariant because the parameters of the system are not changing over time and an
input now will give the same result as the same input later.
i
162
21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS
Applying the Laplace transform on the linear differential equation with null initial
conditions we obtain
an sn Y (s)+an−1 sn−1 Y (s)+· · ·+a0 Y (s) = bm sm U (s)+bm−1 sm−1 U (s)+· · ·+b0 U (s).
The function
Φ(s) =
Y (s)
bm sm + bm−1 sm−1 + · · · + b1 s + b0
=
U (s)
an sn + an−1 sn−1 + · · · + a1 s + a0
is called the system transfer function. That is, the transfer function of a linear
time-invariant system is the ratio of the Laplace transform of its output to the
Laplace transform of its input.
Example 21.4
Consider the mathematical model described by the initial value problem
my 00 + γy 0 + ky = f (t), y(0) = 0, y 0 (0) = 0.
The coefficients m, γ, and k describe the properties of some physical system, and
f (t) is the input to the system. The solution y is the output at time t. Find the
system transfer function.
Solution.
By taking the Laplace transform and using the initial conditions we obtain
(ms2 + γs + k)Y (s) = F (s).
Thus,
Φ(s) =
Y (s)
1
=
2
F (s)
ms + γs + k
(21.1)
Parameter Identification
One of the most useful applications of system transfer functions is for the parameter
identification of a system.
Example 21.5
Consider a spring-mass system governed by
my 00 + γy 0 + ky = f (t), y(0) = 0, y 0 (0) = 0.
(21.2)
Suppose we apply a unit step force f (t) = h(t) to the mass, initially at equilibrium,
and you observe the system respond as
1
1
1
y(t) = − e−t cos t − e−t sin t + .
2
2
2
What are the physical parameters m, γ, and k?
163
Solution.
Start with the model (21.2) with f (t) = h(t) and take the Laplace transform
1
of both sides, then solve to find Y (s) = s(ms2 +γs+k)
. Since f (t) = h(t) we find
F (s) = 1s . Hence
Φ(s) =
1
Y (s)
=
.
F (s)
ms2 + γs + k
On the other hand, for the input f (t) = h(t) the corresponding observed output is
1
1
1
y(t) = − e−t cos t − e−t sin t + .
2
2
2
Hence,
1
1
1
Y (s) =L[− e−t cos t − e−t sin t + ]
2
2
2
1
s+1
1
1
1
=−
−
+
2 (s + 1)2 + 1 2 (s + 1)2 + 1 2s
1
= 2
s(s + 2s + 2)
Thus,
Φ(s) =
Y (s)
1
= 2
.
F (s)
s + 2s + 2
By comparison, we conclude that m = 1, γ = 2, and k = 2
164
21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS
Practice Problems
Problem 21.1
Find the Laplace transform of the periodic function whose graph is shown.
Problem 21.2
Find the Laplace transform of the periodic function whose graph is shown.
Problem 21.3
Find the Laplace transform of the periodic function whose graph is shown.
Problem 21.4
Find the Laplace transform of the periodic function whose graph is shown.
165
Problem 21.5
State the period of the function f (t) and find its Laplace transform where

 sin t, 0 ≤ t < π
f (t + 2π) = f (t), t ≥ 0.
f (t) =

0,
π ≤ t < 2π.
Problem 21.6
State the period of the function f (t) = 1 − e−t , 0 ≤ t < 2, f (t + 2) = f (t), and
find its Laplace transform.
Problem 21.7
Using Example 21.3 find
L−1
s2 − s
e−s
+
.
s3
s(1 − e−s )
Problem 21.8
Consider the initial value problem
ay 00 + by 0 + cy = f (t), y(0) = y 0 (0) = 0, t > 0.
Suppose that the transfer function of this system is given by Φ(s) =
(a) What are the constants a, b, and c?
(b) If f (t) = e−t , determine F (s), Y (s), and y(t).
1
.
2s2 +5s+2
Problem 21.9
Consider the initial value problem
ay 00 + by 0 + cy = f (t), y(0) = y 0 (0) = 0, t > 0.
Suppose that an input f (t) = t, when applied to the above system produces the
output y(t) = 2(e−t − 1) + t(e−t + 1), t ≥ 0.
(a) What is the system transfer function?
(b) What will be the output if the Heaviside unit step function f (t) = h(t) is
applied to the system?
166
21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS
Problem 21.10
Consider the initial value problem
y 00 + y 0 + y = f (t), y(0) = y 0 (0) = 0,
where
f (t) =

 1,

0≤t≤1
f (t + 2) = f (t)
−1, 1 < t < 2.
(a) Determine the system transfer function Φ(s).
(b) Determine Y (s).
Problem 21.11
Consider the initial value problem
y 000 − 4y = et + t, y(0) = y 0 (0) = y 00 (0) = 0.
(a) Determine the system transfer function Φ(s).
(b) Determine Y (s).
Problem 21.12
Consider the initial value problem
y 00 + by 0 + cy = h(t), y(0) = y0 , y 0 (0) = y00 , t > 0.
Suppose that L[y(t)] = Y (s) =
y00 .
s2 +2s+1
.
s3 +3s2 +2s
Determine the constants b, c, y0 , and
22 Convolution Integrals
We start this section with the following problem.
Example 22.1
A spring-mass system with a forcing function f (t) is modeled by the following
initial-value problem
mx00 + kx = f (t), x(0) = x0 , x0 (0) = x00 .
Find solution to this initial value problem using the Laplace transform method.
Solution.
Apply Laplace transform to both sides of the equation to obtain
ms2 X(s) − msx0 − mx00 + kX(s) = F (s).
Solving the above algebraic equation for X(s) we find
X(s) =
msx0
mx00
F (s)
+
+
ms2 + k ms2 + k ms2 + k
(22.1)
=
x00
1 F (s)
sx0
+
+
k
k
2
2
ms + m
s +m
s2 +
k
m
Apply the inverse Laplace transform to obtain
x(t) =L−1 [X(s)]
)
(
(
1 −1
F (s)
s
+ x0 L−1
= L
k
2
2
m
s +m
s +
1
= L−1
m
(
1
F (s) ·
2
s +
)
k
m
)
k
m
r
+ x0 cos
167
(
+ x00 L−1
k
m
!
t+
x00
1
2
s +
r
)
k
m
m
sin
k
r
k
m
!
t
168
Finding L−1 F (s) ·
22 CONVOLUTION INTEGRALS
1
k
s2 + m
,i.e., the inverse Laplace transform of a product, re-
quires the use of the concept of convolution, a topic we discuss in this section
Convolution integrals are useful when finding the inverse Laplace transform of
products H(s) = F (s)G(s). They are defined as follows: The convolution of two
scalar piecewise continuous functions f (t) and g(t) defined for t ≥ 0 is the integral
Z t
f (t − s)g(s)ds.
(f ∗ g)(t) =
0
Example 22.2
Find f ∗ g where f (t) = e−t and g(t) = sin t.
Solution.
Using integration by parts twice we arrive at
Z t
(f ∗ g)(t) =
e−(t−s) sin sds
0
it
1 h −(t−s)
= e
(sin s − cos s)
2
0
e−t 1
=
+ (sin t − cos t)
2
2
Next, we state several properties of convolution product, which resemble those of
ordinary product.
Theorem 22.1
Let f (t), g(t), and k(t) be three piecewise continuous scalar functions defined for
t ≥ 0 and c1 and c2 are arbitrary constants. Then
(i) f ∗ g = g ∗ f (Commutative Law)
(ii) (f ∗ g) ∗ k = f ∗ (g ∗ k) (Associative Law)
(iii) f ∗ (c1 g + c2 k) = c1 f ∗ g + c2 f ∗ k (Distributive Law)
Example 22.3
Express the solution to the initial value problem y 0 + αy = g(t), y(0) = y0 in terms
of a convolution integral.
Solution.
Solving this initial value problem by the method of integrating factor we find
Z t
−αt
y(t) = e y0 +
e−α(t−s) g(s)ds = e−αt y0 + e−αt ∗ g(t)
0
The following theorem, known as the Convolution Theorem, provides a way for
finding the Laplace transform of a convolution integral and also finding the inverse
Laplace transform of a product.
169
Theorem 22.2
If f (t) and g(t) are piecewise continuous for t ≥ 0, and of exponential order a then
L[(f ∗ g)(t)] = L[f (t)]L[g(t)] = F (s)G(s).
Thus, (f ∗ g)(t) = L−1 [F (s)G(s)].
Example 22.4
Use the convolution theorem to find the inverse Laplace transform of
H(s) =
Solution.
Note that
H(s) =
1
.
(s2 + a2 )2
1
2
s + a2
1
2
s + a2
.
1
1
So, in this case we have, F (s) = G(s) = s2 +a
2 so that f (t) = g(t) = a sin (at).
Thus,
Z t
1
1
sin (at − as) sin (as)ds = 3 (sin (at) − at cos (at))
(f ∗ g)(t) = 2
a 0
2a
Convolution integrals are useful in solving initial value problems with forcing functions.
Example 22.5
Solve the initial value problem
4y 00 + y = g(t), y(0) = 3, y 0 (0) = −7
Solution.
Take the Laplace transform of all the terms and plug in the initial conditions to
obtain
4(s2 Y (s) − 3s + 7) + Y (s) = G(s)
or
(4s2 + 1)Y (s) − 12s + 28 = G(s).
Solving for Y (s) we find
Y (s) =
12s − 28
G(s)
1 +
2
4 s +4
4 s2 + 41
1
1
2
=
− 14
+ G(s)
2
1 2
1 2
2
2
2
2
s + 2
s + 2
s + 12
3s
1
2
170
22 CONVOLUTION INTEGRALS
Hence,
Z
s
t
t
1 t
y(t) = 3 cos
− 14 sin
+
sin
g(t − s)ds.
2
2
2 0
2
So, once we decide on a g(t) all we need to do is to evaluate the integral and we’ll
have the solution
Example 22.6
Use Laplace transform to solve for y(t) :
Z
y(t) −
t
e(t−λ) y(λ)dλ = t.
0
Solution.
Note that the given equation reduces to et ∗ y(t) = y(t) − t. Taking Laplace trans(s)
= Y (s)− s12 . Solving for Y (s) we find Y (s) = s2s−1
.
form of both sides we find Ys−1
(s−2)
Using partial fractions decomposition we can write
1
1
− 41
s−1
2
4
=
+
+
.
s2 (s − 2)
s
s2 (s − 2)
Hence,
1
t
1
y(t) = − + + e2t , t ≥ 0
4 2 4
Example 22.7
Use Laplace transform to solve for y(t) :
t ∗ y(t) = t2 (1 − e−t ).
Solution.
Taking Laplace transform of both sides we find
Y (s) =
2
s
−
2s2
.
(s+1)3
Y (s)
s2
=
2
s3
−
2
.
(s+1)3
This implies
Using partial fractions decomposition we can write
s2
1
2
1
=
−
+
.
3
2
(s + 1)
s + 1 (s + 1)
(s + 1)3
Hence,
y(t) = 2 − 2(e−t − 2te−t +
t2
t2 −t
e ) = 2 1 − (1 − 2t + )e−t , t ≥ 0
2
2
171
Example 22.8
Solve the following initial value problem.
Z t
(t − λ)eλ dλ, y(0) = −1.
y0 − y =
0
Solution.
Note that y 0 − y = t ∗ et . Taking Lalplace transform of both sides we find sY −
1
1
1
(−1) − Y = s12 · s−1
. This implies that Y (s) = − s−1
+ s2 (s−1)
2 . Using partial
fractions decomposition we can write
s2 (s
1
2
1
2
1
= + 2−
+
.
2
− 1)
s s
s − 1 (s − 1)2
Thus,
Y (s) = −
1
2
1
2
1
2
1
3
1
+ +
−
+
= + 2−
+
.
s − 1 s s2 s − 1 (s − 1)2
s s
s − 1 (s − 1)2
Finally,
y(t) = 2 + t − 3et + tet , t ≥ 0
172
22 CONVOLUTION INTEGRALS
Practice Problems
Problem 22.1
Consider the functions f (t) = g(t) = h(t), t ≥ 0 where h(t) is the Heaviside unit
step function. Compute f ∗ g in two different ways.
(a) By directly evaluating the integral.
(b) By computing L−1 [F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].
Problem 22.2
Consider the functions f (t) = et and g(t) = e−2t , t ≥ 0. Compute f ∗ g in two
different ways.
(a) By directly evaluating the integral.
(b) By computing L−1 [F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].
Problem 22.3
Consider the functions f (t) = sin t and g(t) = cos t, t ≥ 0. Compute f ∗ g in two
different ways.
(a) By directly evaluating the integral.
(b) By computing L−1 [F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].
Problem 22.4
Compute and graph f ∗ g where f (t) = h(t) and g(t) = t[h(t) − h(t − 2)].
Problem 22.5
Compute and graph f ∗g where f (t) = h(t)−h(t−1) and g(t) = h(t−1)−2h(t−2)].
Problem 22.6
Compute t ∗ t ∗ t.
Problem 22.7
Compute h(t) ∗ e−t ∗ e−2t .
Problem 22.8
Compute t ∗ e−t ∗ et .
Problem 22.9
n f unctions
z
}|
{
Suppose it is known that h(t) ∗ h(t) ∗ · · · ∗ h(t) = Ct8 . Determine the constants C
and the positive integer n.
Problem 22.10
Use Laplace transform to solve for y(t) :
Z t
sin (t − λ)y(λ)dλ = t2 .
0
173
174
Cheat Sheets
CHEAT SHEETS
Answer Key
Section 1
1.1 (a) 16 m/sec (b) 80 m.
1.2(a) 1.5 meters (b) 7 m/sec (c) −9.8 m/sec2 .
1.3 The acceleration due to gravity is y 00 (t) = −50 ft/sec2 . The initial velocity
is v(0) = y 0 (0) = 72 ft/sec. The initial height is y(0) = 40 ft.
1.4 The height of the object at any time t is given by y(t) = −16t2 + 400. The
object hits the ground when y(t) = 0. Solving the equation −16t2 + 400 = 0 we
find t = 5 sec. The velocity of the object at the time of impact is v(5) = y 0 (5) =
−32(5) = −160 ft/sec.
1.5 400 ft.
1.6 v(0) = 160 ft/sec and maximum height is 400 ft.
q
q
√
1.7 The impact time is t = 2yg0 . v( 2yg0 ) = − 2gy0 .
1.8 = π4 .
1.9 (a) Order 2 (b) Order 1 (c) Order 3.
1.10 (a) First order (b) , (c), and (d) second order.
1.11 u is the dependent variable whereas x and y are the independent variables.
1.12 (a) ODE (b) PDE (c) ODE.
181
182
ANSWER KEY
1.13 y(t) =
t3
3
+ C1 t2 + C2 t + C3 .
1.14 The solution is y(t) = 50e3t . The domain is the set of all real numbers,
i.e., (−∞, ∞).
1.15 λ = ±3.
1.16 m = −2 or m = −1.
1.17 Substituting y(t) = y 0 (t) = y 00 (t) = et into the equation we find
2
2
2 t
2 t
00
0
t
y − 2+
y + 1+
y =e − 2 +
e + 1+
e
t
t
t
t
2
2
=et − 2et − et + et + et = 0.
t
t
1.18 Finding the first and the second derivatives of y we obtain
y 0 (t) = −C1 ω sin ωt + C2 ω cos ωt
and
y 00 (t) = −C1 ω 2 cos ωt − C2 ω 2 sin ωt
Substituting this into the equation to obtain
d2 y
+ ω 2 y = − C1 ω 2 cos ωt − C2 ω 2 sin ωt + ω 2 (C1 cos ωt + C2 sin ωt)
dt2
=0.
1.19 y(0) = 2 and k = 4.
1.20 n = 1 or n = 2.
1.21 (b) y(t) = e2t + e−2t (c) y(t) = 2e−2t .
1.22 y(0) = 1, m = −2, y(t) = −t + 2.
1.23 t0 = 0, m = 1, y(t) =
t2
2
− 1.
1.24 Finding the second derivative and substituting into the equation we find
y 00 + 4y = 4e2t + 4e2t = 8e2t 6= 0.
183
Thus, y(t) = e2t is not a solution to the given equation.
1.25 (a) Substituting y and y 0 into the equation we find
−3Ce−3t + 2 + 3[Ce−3t + 2t + 1] = −3Ce−3t + 3Ce−3t + 6t + 5 = 6t + 5.
(b) C = 2.
184
ANSWER KEY
Section 2
2.1 y0 = 3 and p(t) = −2t.
2.2 (a) (−∞, ∞)
(b) (−∞, ∞).
2.3 (a) 3 < t < ∞ (b) −2 < t < 2 (c) −∞ < t < −2.
2.4 (a) r = 3 and C = −1 (b) (−∞, 0) (c) (−∞, ∞).
2.5 The interval of existence is 0 < t < ∞ if y(1) = 1 and −∞ < t < 0 if
y(−1) = 1.
2.6 (a) −5 < t < 0 (b) 0 < t < 4 (c) −∞ < t < −5 (d) 4 < t < ∞.
2.7 (a) 1 < t < 3 (b) No such solution since t > 0 (c) 3 < t < ∞.
2.8 (a) If p(t) and g(t) are continuous functions in the open interval I = (a, b) and
t0 a point inside I then the IVP
y 0 + p(t)y = g(t), y(t0 ) = y0
has a unique solution y(t) defined on I.
3
(b) Since p(t) = t2 and g(t) = et , the IVP has a unique solution.
2.9 (a) Non-linear (b) Linear and homogeneous (c) Linear and non-homogeneous (d)
Linear and homogeneous.
2.10 0 < t < 5.
2.11 α = ln 2 and y0 = 8.
2.12(a) (iv)
(b) (i)
(c) (iii)
(d) (ii).
2.13 A non-linear first order ordinary differential equation.
2.14 We have y 0 =
t+y
t
= 1 + yt . Thus,
1
y0 − y = 1
t
This is a first-order linear non-homogeneous equation.
185
Section 3
2
3.1 y(t) = e1−t
t
3.2 y(t) = 2e1−e .
−
3.3 (a)z 0 + p(t)z = 0, z(t0 ) = y(t0 ) − α (b) y(t) = (y0 − α)e
2
3.4 y(t) = e−t + 3.
3.5 α = 2 and y0 = 0.25.
3.6 (a) n = 3 and y0 = 1 (b) y(−1) =
√
4
e−1 .
3.7 p(t) = 1t .
3.8 p(t) = − 3t .
3.9 y(t) = t3 .
2
3.10 y(t) = Cet .
3.11 f (t) = 4e2t .
3.12 y(t) = Ce2t + C 0 .
2
3.13 y = 12 (1 − e−t ).
3.14 y(t) =
3t−1
9
+ e−2t + Ce−3t .
3.15 y(t) = 3 sin t +
3.16 y(t) =
3 cos t
t
1
13 (3 sin (3t)
+
C
t.
+ 2 cos (3t)) + Ce−2t .
3.17 α = −2.
3.18 p(t) = 2 and g(t) = 2t + 3.
3.19 y0 = −1 and g(t) = 2et + cos t + sin t.
Rt
t0
p(s)ds
+ α.
186
ANSWER KEY
3.20 limt→∞ y(t) = 1.
3.21
y(t) =
3.22
y(t) =
1 + 2ecos t−1
if 0 ≤ t ≤ π
−1 + 2 1e + e ecos t if π < t ≤ 2π.
3.23
y(t) =
3.24
2t + 1 if 0 ≤ t ≤ 1
t + 2t if 1 < t ≤ 2.
1 + 2e−t if 0 ≤ t ≤ 1
(2 + e)e−t
if t > 1.
 −(t2 −t)
if 0 ≤ t ≤ 1
 3e
y(t) =
3
if 1 < t ≤ 3

t
if 3 < t ≤ 4.
The graph of y(t) is shown below
It follows that limt→∞ y(t) = ∞. The function y(t) is not differentiable at t = 1
and t = 3 on the domain t > 0.
3.25 y(t) = (−tet + et )−1 .
187
Section 4
4.1 (a) y 0 = 13 (1 − 2t cos y) = f (t, y),
(b) D = {(t, y) : −∞ < t < ∞, − ∞ < y < ∞}
(c) R = {(t, y) : −∞ < t < ∞, − ∞ < y < ∞}.
1
4.2 (a) y 0 = 3t
(1 − 2 cos y) = f (t, y)
(b) D = {(t, y) : −∞ < t < 0, 0 < t < ∞, − ∞ < y < ∞}
(c) R = {(t, y) : 0 < t < ∞, − ∞ < y < ∞}.
2t
4.3 (a) y 0 = − 1+y
3 = f (t, y)
(b) D = {(t, y) : −∞ < t < ∞, − ∞ < y < −1, − 1 < y < ∞}
(c) R = {(t, y) : −∞ < t < ∞, − 1 < y < ∞}.
2
−y
4.4 (a) y 0 = t y−e
2 −9 = f (t, y)
(b) D = {(t, y) : −∞ < t < ∞, − ∞ < y < −3, − 3 < y < 3, 3 < y < ∞}
(c) D = {(t, y) : −∞ < t < ∞, − 3 < y < 3}.
t
4.5 (a) y 0 = 2+tan
cos y = f (t, y)
(b) D = {(t, y) : t 6= (2n + 1) π2 , y 6= (2m + 1) π2 , where n and m are integers}
(c) R = {(t, y) : − π2 < t < π2 , − π2 < y < π2 }.
4.6 y 0 =
1
t(t−4)(y+1)(y−2) ,
4.7 Since f (t, y) =
region
y
√
t
y(2) = 0.
and fy (t, y) =
1
√
,
t
both functions are continuous in the
D = {(t, y) : 0 < t < ∞, − ∞ < y < ∞}
Since (0, 2) is not in D, Theorem 4.1 can not be applied in this case.
4.8 The function f (t, y) = t−y
t+y is continuous everywhere except along the line
t + y = 0. (a) and (b): Since both (0, 0) and (1, −1) lie on this line, we cannot
conclude existence from Theorem 4.1.
(c) The point (−1, −1) is not on that line so we can find a small rectangle around
this point where Theorem 4.1 guarantees the existence of a solution. Furthermore,
2t
since fy (t, y) = − (t+y)
2 is continuous at (−1, −1), the solution is unique.
4.9 The equation is of the form y 0 = f (t, y) = yt + 2. The function f is continuous
outside the line t = 0. The initial value point is (0,1), so there is no rectangle
containing it in which f is continuous, and the conditions of Theorem 4.1 are not
satisfied.
188
ANSWER KEY
4.10 The equation is of the form y 0 = f (t, y) = y sin y + t. the function f is continuous in the whole plane, and so is its partial derivative fy (t, y) = sin y + y cos y. In
particular, any rectangle around the initial value point will satisfy the conditions
of Theorem 4.1.
189
Section 5
2
5.1 y 3 = 32 et + C.
t2
5.2 y(t) = Ce 2 −2t .
5.3 y(t) = Ct2 + 4.
5.4 y(t) =
2Ce4t
1+Ce4t
5.5 y(t) =
p
5 − 4 cos (2t).
and y(t) ≡ 2.
p
5.6 y(t) = − (−2 cos t + 4).
5.7 y(t) =
5.8 y(t) =
√
√
−2t + 3 − 1.
2
.
−4t2 +1
5.9 y(t) = tan t −
π
2
, −
π
2
< t < π2 .
2
5.10 y(t) =
3−e−t
.
3+e−t2
5.11 y 2 + cos y + cos t +
t2
2
= 2.
5.12 y0 = 21 , α = 21 , n = 3.
5.13 3y 2 y 0 + (cos y)y 0 + 2t = 0, y(2) = 0.
5.14 y0 = 18 .
5.15 y sin y − t3 + 1 = 0.
5.16 y(t) =
Ce2t
.
(t+1)2
5.17 tan y =
t
2
+ 14 sin 2t + 1.
5.18 y(t) = − ln(2 − et ), t < ln 2.
3
5.19 (a) e−y = − t3 −
1
t
+ C (b) e−y +
t3
3
+
1
t
= e−1 + 43 .
190
ANSWER KEY
Section 6
6.1
∂f
∂t
= ye−ty ,
6.2
∂f
∂t
=
6.3
df
dt
= 24 sin2 t − 4 sin t − 12.
1
t
+
∂f
∂y
= ln y + 1 + te−ty .
∂f
∂y
2t
y−5 ,
1
y
=
−
t2 +1
.
(y−5)2
√
6.4 y(t) = − 4t − t3 + 6.
6.5 y(t) = tan t3 + t +
π
4
.
6.6 The given differential equation is not exact.
6.7 et+y + t3 + y 2 = 1.
6.8 y sin (t + y) +
t2
2
+ ty +
y2
2
= 0.
6.9 m = 4, n = 3, α = 43 .
6.10 N (t, y) =
R
2y sin tdt = −2y cos t + h(y).
6.11 y0 = ±2, M (t, y) = 3t2 + et , N (t, y) = t3 + 2y.
6.12 y(0) = −2, a = 1, b = 2.
6.13 The differential equation is exact.
6.14 t2 + 3t + y 2 − 2y = C.
6.15 B = 1,
1 2ty
2e
+
t2
2
= C.
6.16 ty 2 − y 2 ey + 2yey − 2ey = C.
∂M
∂y (t, y)
R
−
p(t)dt
Ce
.
6.17 (a)
= p(t) 6= 0 and
6.18 y(t) = t ln t + 7t.
6.19 ty 3 − t2 y = C.
∂N
∂t (t, y)
= 0 (b) y(t) = e−
R
p(t)dt
R
e
R
p(t)dt g(t)dt+
191
6.20 (a) 2t3 y 2 + 4ty − 8t + 4y = 6 (b) 6t − 4ty + 8t2 + 5y 2 + 2y = 30.
192
ANSWER KEY
Section 7
7.1 y 2 = −t2 + Ct3 .
2
2
7.2 y 4 = 2e−t + Ce−2t .
7.3 y(t) =
1
t(4.5−t) .
7.4 y(t) =
2t
.
t2 +1
7.5 y(t) = (1 + e−t )−1 .
7.6 y(t) = (t +
7.7 y(t) = (et
2
1
3
Rt
0
1
+ Ce3t )− 3 and y(t) ≡ 0.
2
2
e−s ds + Cet )−1 + t.
7.8 z 0 − z = 1.
1
7.9 (a) z 0 + 2z = 2σ (b) y(t) = ( σ + Ce−2t )− 2 and y(t) ≡ 0.
y
2 1+ y 2 .
7.10 (a) z 0 = f (z)−z
(b)
2
arctan
=
ln
t
t
t
t
1
7.11 y(t) = (−3tet + Cet )− 3 and y(t) ≡ 0.
1
7.12 y(t) = (2t + Ct2 )− 2 and y(t) ≡ 0.
7.13 y(t) = (−t2 ln t + t2 + Ct)−1 and y(t) ≡ 0.
7.14 y(t) = (− 31 + Ce−3t )−1 + 2.
7.15 y(t) = (− 4t + Ct−3 )−1 + 2t .
193
Section 8
8.1 (a) We have
(b) We have
8.2 (A) y 0 = sin t
(B) y 0 = y(2 − y)
(c) y 0 = −t
(D) y 0 = 1 − y
194
8.3 (a) No
ANSWER KEY
(b) No
(c) Yes
(d) Yes
8.4 (a) y(t) ≡ 1, y(t) ≡ 2 (b) y(t) ≡ 1, y ≡ 2 (c) y ≡ 1, y ≡ 2, y ≡ 3
8.5 One answer is the differential equation: y 0 = −(y − 1)2
8.6 y 0 = C, where C > 0.
8.7 One answer is the DE y 0 = sin (2πy)
8.8 An answer is y 0 = y(2 − y)
8.9 (a) y = 1 stable, y = −1 unstable (b) y = −1 is neither. This is a semistable equilibrium.
8 .10 The equilibrium solution at y = 1 is asymptotically stable where as the
equilibrium solution at y = 0 is unstable.
8.11
8.12 The direction field is given below.
195
The equilibrium solution y = 1 is unstable; y = 0 is semi-stable; y = −1 is asymptotically stable.
8.13 The direction field is given below.
The equilibrium solution y = 4 is unstable while y = 0 is asymptotically stable.
If y(0) = 5 then limt→∞ y(t) = ∞. If y(0) = 4 then limt→∞ y(t) = 4. If y(0) = 3
then limt→∞ y(t) = 0
8.14 y(t) = 0 and y(t) = π are two equilibrium solutions. According to the
direction field shown below we conclude that
lim y(t) = π
t→∞
196
ANSWER KEY
197
Section 9
9.1 (a) Integrating y 0 = 2t − 1 we find y(t) = t2 − t + C. Imposing the initial
condition y(1) = 0, we obtain y(t) = t2 − t.
(b) yn+1 = yn + h2 [(2tn − 1) + (2tn+1 − 1)]. (c) yn+1 = yn + h[2(tn + h2 ) − 1]. (d)
n
0
1
2
3
tn
1.0000
1.1000
1.2000
1.3000
yn
0
0.1100
0.2400
0.3900
n
0
1
2
3
tn
1.0000
1.1000
1.2000
1.3000
yn
0
0.1100
0.2400
0.3900
n
0
1
2
3
tn
1.0000
1.1000
1.2000
1.3000
yn
0
0.1100
0.2400
0.3900
(e)
(f)
9.2 (a) Integrating y 0 = −y and imposing the initial condition we find y(t) = e−t .
(b) Heun’s method takes the form yn+1 = (1 − h + 0.5h2 )yn .
(c) The modified Euler’s method takes the form yn+1 = (1 − h + 0.5h2 )yn .
(d)
n
0
1
2
3
(e)
tn
0.0000
0.1000
0.2000
0.3000
yn
1.0000
0.9050
0.8190
0.7412
198
ANSWER KEY
n
0
1
2
3
tn
0.0000
0.1000
0.2000
0.3000
yn
1.0000
0.9050
0.8190
0.7412
n
0
1
2
3
tn
0.0000
0.1000
0.2000
0.3000
yn
1.0000
0.9048
0.8187
0.7408
(f)
9.3 (a) Solving the separable equation y 2 y 0 + t = 0 we find y 3 = −1.5t2 + C.
1
Imposing the initial condition y(0) = 1, we obtain y(t) = (1 − 1.5t2 ) 3 .
(b) yn+1 = yn + h2 [−tn yn−2 − tn+1 (yn − htn yn−2 )−2 ].
(c) yn+1 = yn − h(tn + 0.5h)(yn − 0.5htn yn−2 )−2 .
(d)
n
0
1
2
3
tn
0.0000
0.1000
0.2000
0.3000
yn
1.0000
0.9950
0.9796
0.9529
n
0
1
2
3
tn
0.0000
0.1000
0.2000
0.3000
yn
1.0000
0.9950
0.9797
0.9531
n
0
1
2
3
tn
0.0000
0.1000
0.2000
0.3000
yn
1.0000
0.9950
0.9796
0.9528
(e)
(f)
9.4 (a) y(t) = tan t −
(b)
π
4
.
199
n
0
1
2
3
4
tn
0.0000
0.0500
0.1000
0.1500
0.2000
Euler
−1.0000
−0.9000
−0.8095
−0.7267
−0.6503
Heun
Modified Euler
−1.0000 −1.0000
−0.9047 −0.9049
−0.8177 −0.8179
−0.7375 −0.7378
−0.6630 −0.6634
The errors at t = 1 are, respectively,
| tan 1 − π4 +0.6503| = 0.8683, | tan 1 − π4 +
0.6630| = 0.8809, and | tan 1 − π4 + 0.6634| = 0.8814.
9.5 (a) Solving the equation y 0 + 2y = 4 by the method of integrating factor,
we find y(t) = 2 + Ce−2t . Imposing the initial condition, we obtain C = 1. Hence,
y(t) = 2 + e−2t .
(b) Since f (t, y) = 4 − 2y, we have:
Euler’s method: yn+1 = yn + h(4 − 2yn ).
Heuns’ Method: yn+1 = yn + 0.5h[8 − 4yn − 2h(4 − 2yn )].
Modified Euler’s Method: yn+1 = yn + h[4 − 2yn − h(4 − 2yn )].
n
0
1
2
3
4
tn
0.0000
0.0500
0.1000
0.1500
0.2000
Euler
3.0000
2.9000
2.8100
2.7290
2.6561
Heun
3.0000
2.9050
2.8190
2.7412
2.6708
Modified Euler
3.0000
2.9050
2.8190
2.7412
2.6708
The errors at t = 1 are, respectively, |3 − 2.6561| = 0.3439, |3 − 2.6708| = 0.3292,
and |3 − 2.6708| = 0.3292
9.6 Since tn = 2 + nh, h = 0.02, n = 0, 1, · · · , 49, it follows that t0 = 2 and
N − 1 = 49. Thus, N = 50 and T = tN − t0 = 2 + N h − 2 = 50(0.02) = 1. From
the form of the iteration, it must be Euler’s method. Therefore f (t, y) = y + t2 y 3 .
9.7 Since tn = nh, h = 0.01, n = 0, 1, · · · , 199, it follows that t0 = 0 and
N − 1 = 199. Thus, N = 200 and T = tN − t0 = N h = 200(0.01) = 2. From
the form of the iteration, it must be the modified Euler’s method. Therefore,
f (t, y) = t sin2 y.
9.8 Since tn = 1 + nh, h = 0.05, n = 0, 1, · · · , 99, it follows that t0 = 1 and
N − 1 = 99. Thus, N = 100 and T = tN − t0 = 1 + N h − 1 = 100(0.05) = 5. From
the form of the iteration, it must be Heun’s method. Therefore f (t, y) = ty 2 + 1.
200
ANSWER KEY
Section 10
10.1 π2 < t < 3π
2 .
10.2 −∞ < t < −1.
10.3 0 < t < 3.
10.4 −∞ < t < 0.
10.5 0 < t < π.
π
π
10.6 e− 2 < t < e 2 .
10.7 y 00 + y 0 + y = 0, y(0) = 0, y 0 (0) = 1.
1 0
10.8 y 00 + t−3
y + y = 1, y(4) = 0, y 0 (4) = −1.
10.9 0 < t < 3.
10.10 Let Y (t) = y(−t). Then Y 00 + t2 Y = y 00 + t2 y = 0, Y (0) = Y 0 (0) = 0 so that
Y (t) is a solution to the given initial-value problem. By the existence and uniqueness theorem we must have Y (t) = y(t), i.e., y(−t) = y(t) for all real numbers t.
10.11 If p(t), q(t), and g(t) are continuous functions over an interval a < t < b
containing t0 then the initial value problem
y 00 + p(t)y 0 + q(t)y = g(t), y(t0 ) = y0 , y 0 (t0 ) = y00
has a unique solution in the interval a < t < b.
10.12 We have y 0 (t) = −2t sin (t2 ) and y 00 (t) = −2 sin (t2 ) − 4t2 cos (t2 ). Hence,
ty 00 − y 0 + 4t3 y =t(−2 sin (t2 ) − 4t2 cos (t2 )) − (−2 sin (t2 )) + 4t3 cos (t2 )
= − 2t sin (t2 ) − 4t3 cos (t2 ) + 2 sin (t2 ) + 4t3 cos (t2 ) = 0.
10.13 −1 < t < 4.
10.14 0 < t < π.
10.15
L(αy1 + βy2 ) =(αy1 + βy2 )00 + p(t)(αy1 + βy2 )0 + q(t)(αy1 + βy2 )
=αy100 + βy200 + p(t)αy10 + p(t)βy20 + q(t)αy1 + q(t)βy2
=α(y100 + p(t)y10 + q(t)y2 ) + β(y200 + p(t)y20 + q(t)y2 )
=αL(y1 ) + βL(y2 ).
201
Section 11
11.1 (a) Yes (b) {y1 , y2 } is a fundamental set (c) y = e−2t .
11.2 (a) y1 is not a solution.
11.3 (a) Yes (b) {y1 , y2 } is a fundamental set (c) y(t) = 2e2t − 4te2t .
11.4 (a) Yes (b) {y1 , y2 } is a fundamental set (c) y(t) = 18 ln t − 9 ln (3t), t > 0.
11.5 (a) Yes (b) {y1 , y2 } is a fundamental set (c) y(t) = 12 (t−1 − t3 ), t > 0.
11.6 (a) Yes (b) {y1 , y2 } is a fundamental set (c) y(t) = −t + 5.
11.7 (a) y1 is not a solution.
11.8 (a) t2 y 00 − ty 0 + y = t2 t−1 − t(3 + ln (3t)) + 2t + t ln (3t) = 0 (b) c1 = 2 + ln 3
and c2 = 1.
11.9 α = 0 and β = −9.
1
−t
and q(t) = t−1
(b) (−∞, 1) ∪ (1, ∞) (c) W (y1 (t), y2 (t)) =
11.10 (a)p(t) = t−1
et (t − 1) The Wronskian is nonzero for all t 6= 1 (d) Yes.
11.11 W (y1 (2), y2 (2)) = 4e−1.5 .
11.12 α = 1 and β = −6.
11.13 p(t) = 2t.
11.14 Suppose for example that both functions have a same maximum at t0 .
Then y10 (t0 ) = y20 (t0 ) = 0. But
W (y1 (t0 ), y2 (t0 )) = y1 (t0 )y20 (t0 ) − y10 (t0 )y2 (t0 ) = 0
Thus, {y1 , y2 } is not a fundamental set.
11.15 W (y1 (t), y2 (t)) = W (y1 (t0 ), y2 (t0 ))
11.16 y2 (t) = tet .
1
t
−
1
t0
.
202
ANSWER KEY
11.17 {y1 , y2 } is a fundamental set.
11.18 Suppose that t3 and t4 are both solutions. Since W (t) = t6 we find
W (1) = 1 and so {y1 , y2 } is a fundamental set. By Abel’s Theorem, W (t) 6= 0 for
all −∞ < t < ∞. But W (0) = 0, a contradiction. Hence, t3 and t4 can’t be both
solutions for the differential equation for −∞ < t < ∞.
11.19 p(t) = − t22t+1 .
203
Section 12
12.1 y(t) = et + 2e−2t , limt→−∞ y(t) = ∞, limt→∞ y(t) = ∞.
12.2 y(t) = −2et + e3t , limt→−∞ y(t) = ∞, limt→∞ y(t) = ∞.
12.3 y(t) = e−t , limt→−∞ y(t) = ∞, limt→∞ y(t) = 0.
12.4 y(t) = 2e−2t − e−3t , limt→−∞ y(t) = −∞, limt→∞ y(t) = 0.
12.5 y(t) = 0, limt→−∞ y(t) = 0, limt→∞ y(t) = 0.
12.6 y(t) = 3, limt→−∞ y(t) = 3, limt→∞ y(t) = 3.
√
√
√
√
12.7 y(t) = − 2e(−2− 2)t + 2e(−2+ 2)t , limt→−∞ y(t) = −∞, limt→∞ y(t) = 0.
√
−
12.8 y(t) = −2e
12.9 y(t) =
R
2
t
2
, limt→−∞ y(t) = −∞, limt→∞ y(t) = 0.
u(t)dt = C1 e2t + C2 e3t + C3 .
k
12.10 (a) x(t) = c1 +c2 e− m t (b) x(t) = x0 +
x0 + m
k v0 .
12.11 y 00 − y 0 − 2y = 0.
12.12 y(t) = c1 e−t + c2 e4t .
12.13 y(t) = c1 et + c2 e−5t .
12.14 y(t) = c1 et + c2 e
−t
3
.
12.15 y(t) = − 15 (3et + 2e−4t ).
t
12.16 y(t) = 2e 2 .
m
k v0
k
−mt
−m
(c) limt→∞ x(t) =
k v0 e
204
ANSWER KEY
Section 13
t
t
t
t
13.1 (a) y(t) = c1 e 3 +c2 te 3 (b) y(t) = e 3 −1 (1−t) (c) limt→∞ y(t) = − limt→∞ e 3 −1 (t−
1) = −∞.
2t
2t
2t
13.2 (a) y(t) = c1 e− 5 + c2 te− 5 (b) y(t) = e− 5 (t − 1) (c) limt→−∞ y(t) = −∞
and limt→∞ y(t) = 0.
13.3 (a)y(t) = c1 e2t + c2 te2t (b) y(t) = 4e2t−2 (2t − 3) (c) limt→−∞ y(t) = 0
and limt→∞ y(t) = ∞.
√
√
√
13.4 (a) y(t) = c1 e− 2t +c2 te− 2t (b) y(t) = e−
limt→−∞ y(t) = −∞ and limt→∞ y(t) = 0.
13.5 (a) y(t) = c1 ert + c2 tert (b) y(t) = e
and limt→∞ y(t) = 0.
13.6 y(t) = c1 e3t + c2 te3t .
t
t
13.7 y(t) = c1 e 2 + c2 te 2 .
t
t
13.8 y(t) = 2e− 2 + te− 2 .
− √t
3
2t +
√
2te−
√
2t
= e−
√
2t (1+
√
2t) (c)
√
(5t + 2 3) (c) limt→−∞ y(t) = −∞
205
Section 14
14.1 Adding z and z we find 2α = z + z. Hence, α = 21 (z + z). Next, subtracting
1
z from z we find 2iβ = z − z. Therefore, β = 2i
(z − z).
√
π
14.2 1. 2ei 3 = 2 cos ( π3 ) + 2i sin ( π3 ) = 1 + i 3
3π
2. (2√− i)ei 2 = −1 − 2i
√
π
3. ( 2ei 6 )4 = −2 + 2i 3.
√
√
√
14.3 1. 2ei 2t = 2 cos 2t + 2i sin 2t
2. − 12 e2t+i(t+π) = 21 e2t cos t + 12 ie2t sin t
√
√
√
3. ( 3e(1+i)t )3 = 3 3e3t cos (3t) + 3 3ie3t sin (3t).
14.4 (a) r1 = −1 − i and r2 = −1 + i
(b) y(t) = e−t (c1 cos t + c2 sin t)
(c) y(t) = 3e−t cos t + 2e−t sin t.
14.5 (a) r1 = 21 (1 − i) and r2 = 12 (1 + i)
t
(b) y(t) = e 2 (c1 cos 2t + c2 sin 2t )
1
(c) y(t) = −e 2 (t+π) (3 cos 2t + sin 2t ).
14.6 (a) r1 = −2 − i and r2 = −2 + i
(b) y(t) = e−2t (c1 cos t + c2 sin t)
(c) y(t) = eπ−2t (cos t + 12 sin t).
14.7 (a) r1 = −2πi and r2 = 2πi
(b) y(t) = c1 cos 2πt + c2 sin 2πt
(c) y(t) = 2 cos 2πt + (2π)−1 sin 2πt.
14.8 (a) r1 = − π3 i and r2 = π3 i
(b) y(t) = c1 cos π3 t + c2 sin π3 t
(c) y(t) = −2 cos π3 t + 3 sin π3 t.
14.9 a = 0, b = 4, y0 = 2, y00 = −2.
14.10 a = −2, b = 5, y0 =
1
2
√
−
√
14.11 y(t) = 2 cos t − π4 .
The graph of y(t) is given below.
3
2 ,
y00 = − 12 −
√
3 3
2 .
206
14.12 y(t) = 2et cos t − π3 .
The graph of y(t) is given below.
√
14.13 y(t) = 2e−2t cos 2t − 7π
4 .
The graph of y(t) is given below.
ANSWER KEY
207
14.14 y(t) = c1 cos 3t + c2 sin 3t.
√
14.15 y(t) = 5 5 cos 7t + arctan
11
2
.
14.16 y 00 − 4y 0 + 13y = 0.
14.17 y 00 − 10y 0 + 29y = 0.
14.18 y 00 + 4y = 0, y(0) = 3, y 0 (0) = −8.
1
14.19 Since x = ln t we find dx
dt = t . But
dy
d2 y
d2 y
dy
1
− t12 dx
+ t12 dx
− dx
. Hence,
2 = t2
dx2
dy
dt
=
dy dx
dx dt
=
1 dy
t dx .
0 =t2 y 00 + αty 0 + βy
2
1 d y
dy
1 dy
2
=t
−
+ αt
+ βy
t2 dx2 dx
t dx
d2 y
dy
+ βy
= 2 + (α − 1)
dx
dx
14.20 y(t) = c1 cos (ln t) + c2 sin (ln t).
Moreover,
d2 y
dt2
=
208
ANSWER KEY
Section 15
15.1 Since p(t) = t and q(t) = t2 + 2 are polynomials, they are analytic functions everywhere. Thus, every real number is an ordinary point.
t
15.2 We have p(t) = t2 −1
and q(t) = t(t21−1) . Thus, the points 1, −1, and 0
are singular points of the equation, any other real number is an ordinary point of
the equation.
15.3 We have p(t) = t23t−4 and q(t) = t21−4 . Thus, the singular points are t = ±2
and any other point is an ordinary point.
15.4 R = 2.
15.5 y(t) = a0 1 − t2 + 14 t4 −
15.6 y(t) = a0 1 +
1 4
12 t
+
1 6
60 t
1 6
90 t
+ · · · + a1 t + 16 t3 +
15.7 y(t) = a0 1 − 12 t2 − 16 t3 +
15.8
P
n=2 [(n
+ · · · + a1 t − 12 t3 +
1 4
12 t
+
3 5
40 t
3 5
40 t
3 5
40 t
+
−
1 7
1680 t
13 7
1008 t
+ ··· .
+ ··· .
+ · · · +a1 t + 12 t2 − 18 t4 −
1 5
40 t
+ ··· .
− 1)an−1 + n2 an ](t − 2)n .
15.9 y(t) = a0 1 − 12 t2 + 18 t4 −
1 6
48 t
+ · · · + a1 t − 13 t3 +
1 5
15 t
−
1 7
105 t
+ ··· .
15.10 y(t) = a0 1 + t2 + 23 t3 + 23 t4 + 23 t5 + · · · + a1 t + 13 t3 + 31 t4 + 13 t5 + · · · .
209
Section 16
16.1 (a) yp0 = − 23 e2t , yp00 = − 34 e2t , and yp00 − 2yp0 − 3yp = − 43 e2t + 34 e2t + e2t = e2t .
(b) yh (t) = c1 e−t + c2 e3t
(c) The general solution to the differential equation is y(t) = c1 e−t + c2 e3t − 13 e2t
and the solution to the IVP is y(t) = 56 e−t + 21 e3t − 13 e2t .
16.2 (a) yp0 = yp00 = 0 and yp00 − yp0 − 2yp = 0 − 0 − 2(−5) = 10.
(b) yh (t) = c1 e−t + c2 e2t .
(c) The general solution to the differential equation is y(t) = c1 e−t + c2 e2t − 5 and
the solution to the IVP is y(t) = 3e−(t+1) + 2e2t+2 − 5.
16.3 (a) yp0 = −2e−t + 2te−t , yp00 = 4e−t − 2te−t and yp00 + yp0 = 4e−t − 2te−t −
2e−t + 2te−t = 2e−t .
(b) yh (t) = c1 + c2 e−t .
(c) The general solution to the differential equation is y(t) = c1 + c2 e−t − 2te−t
and the solution to the IVP is y(t) = 6 − 4e−t − 2te−t .
16.4 (a) yp0 = yp00 = 2et−π and yp00 + 4yp0 = 2et−π + 8et−π = 10et−π .
(b) yh (t) = c1 cos 2t + c2 sin 2t.
(c) The general solution to the differential equation is y(t) = c1 cos 2t + c2 sin 2t +
2et−π and the solution to the IVP is y(t) = − sin 2t + 2et−π .
16.5 (a) yp0 = −2 sin t + cos t, yp00 = −2 cos t − sin t, and yp00 − 2yp0 + 2yp =
−2 cos t − sin t + 4 sin t − 2 cos t + 4 cos t + 2 sin t = 5 sin t.
(b) yh (t) = et (c1 cos t + c2 sin t).
(c) The general solution to the differential equation is y(t) = et (c1 cos t + c2 sin t) +
π
2 cos t + sin t and the solution to the IVP is y(t) = −2et− 2 cos t + 2 cos t + sin t.
16.6 (a) yp0 = 2t + 4 − sin t, yp00 = 2 − cos t, and yp00 − 2yp0 + yp = 2 − cos t −
4t − 8 + 2 sin t + t2 + 4t + 10 + cos t = t2 + 4 + 2 sin t.
(b) yh (t) = c1 et + c2 tet .
(c) The general solution to the differential equation is y(t) = c1 et + c2 tet + t2 +
4t + 10 + cos t and the solution to the IVP y(t) = −10et + 9tet + t2 + 4t + 10 + cos t.
16.7 u(t) = 21 u1 (t) + 32 u3 (t).
16.8 u(t) = 14 u1 (t) + 41 u2 (t) +
3
1
16.9 g(t) = − 14 t− 2 − 6 − t− 2 .
1
12 u3 (t).
210
1
16.10 g(t) = − (1+t)
2 +
ANSWER KEY
1
1+t .
16.11 g(t) = t.
16.12 α = −2, β = 1, g(t) = 2et .
16.13 α = 0, β = 4, g(t) = 3 sin t − 4.
16.14 y(t) = c1 cos 2t + c2 sin 2t +
et
5.
16.15 y(t) = c1 cos 3t + c2 sin 3t + c3 t + c4 + t4 .
211
Section 17
17.1 y(t) = c1 cos t + c2 sin t + ln | cos t| cos t + t sin t.
17.2 y(t) = c1 et + c2 e−t + 21 tet .
17.3 y(t) = c1 cos 2t + c2 sin 2t −
2
17.4 yp (t) = ( t2 ln t −
17.5 y =
et
2
64 (8t
1
8
+ 14 t2 + 2t sin 2t.
3t2 −t
4 )e .
− 4t + 15) − 41 e−3t .
√
17.6 (a) Usually
the first thing to do would be to check that y1 (t) = e t and
√
y2 (t) = e− t really are solutions to the equation. However, the question says that
this can be assumed and so we move on to the next step, which is to check that
the Wronskian of the two solutions is non-zero on (0, ∞). We have
y10 =
and so
√
1
t
√
e
2 t
1 −
and y20 = − 2√
e
t
√
t
1
1
1
W (t) = y1 y20 − y10 y2 = − √ − √ = − √
2 t 2 t
t
√
√
This is indeed non-zero and so {e t , e− t } is a fundamental set for the homogeneous equation.
√ √
(b) y = (t − t)e t .
17.7 y(t) = c1 cos t + c2 sin t − 21 t cos t.
17.8 (a) r1 = 1 and r2 = −3.
(b) y1 (t) = t and y2 (t) = t−3 .
1 −3
(c) yp (t) = − 16
t − 14 t−3 ln t.
17.9 y(t) = c1 cos t + c2 sin t + cos2 t − 31 cos4 t + 13 sin4 t.
212
ANSWER KEY
Section 18
18.1
R∞
18.2
R∞
18.3
R∞
0
0
0
1
dt
1+t2
= π2 .
t
dt
1+t2
= ∞.
e−t cos (e−t )dt = sin 1.
1
s−3 ,
18.4 L[e3t ] =
18.5 L[t − 5] =
1
s2
s > 3.
− 5s , s > 0.
2
18.6 f (t) = e(t−1) does not have a Laplace transform.
18.7 F (s) =
4
s
18.8 L[f (t)] =
+
2
s
− 2s +
e−s
,
s2
1
s2
=
4
s
−
4
s2
+
2
,
s3
s > 0.
s > 0.
−2s
18.9 L[f (t)] = − e s +
1
(e−s
s2
− e−2s ), s 6= 0.
−st
18.10 Let u0 = e−st and v = tn . Then u = − e s
Z
tn e−st dt = −
tn e−st n
+
s
s
Z
and v 0 = ntn−1 . Hence,
tn−1 e−st dt, s > 0.
18.11 (a) limt→0 tn e−st = 0 (b) limt→∞ tn e−st = 0.
18.12 (a) Using the two previous problems we find
L[tn ] = lim
T
)
( Z T
n e−st T
n
t
+
tn e−st dt = lim −
tn−1 e−st dt
T →∞
s
s
0
0
Z
T
Z
T →∞ 0
n
= lim
s T →∞
0
tn−1 e−st dt =
n n−1
L[t
], s > 0
s
213
(b) We have
1
s2
2
2
L[t2 ] = L[t] = 3
s
s
3 2
6
3
L[t ] = L[t ] = 4
s
s
24
4
L[t4 ] = L[t3 ] = 5
s
s
5 4
120
5
L[t ] = L[t ] = 5
s
s
L[t] =
(c) By induction, one can easily show that for n = 0, 1, 2, · · ·
L[tn ] =
18.13 L[cos ωt] =
s
,
s2 +ω 2
s > 0.
18.14 L[sin ωt] =
ω
,
s2 +ω 2
s > 0.
18.15 L[cos ω(t − 2)] =
18.16 L[e3t sin t] =
n!
sn+1
s cos 2ω+ω sin 2ω
,
s2 +ω 2
1
,
(s−3)2 +1
, s > 0.
s > 0.
s > 3.
18.17 (a) Neither (b) The graph of the function does not show that it can be
bounded by exponential functions. Hence, no such numbers a and M.
t2
18.18 (a) f (t) = e2te +1 is continuous on 0 ≤ t < ∞ (b) f (t) is not of exponential order at infinity.
3
18.19 L−1 s−2
= 3e2t , t ≥ 0.
1
18.20 L−1 − s22 + s+1
= −2t + e−t , t ≥ 0.
214
ANSWER KEY
Section 19
19.1 L[2et + 5] = 2L[et ] + 5L[1] =
2
s−1
+ 5s , s > 1.
−s
e
19.2 L[e3t−3 h(t − 1)] = s−3
, s > 3.
s
19.3 L[sin2 ωt] = 12 1s − s2 +4ω
, s > 0.
2
19.4 L[sin 3t cos 3t] =
19.5 L[e2t cos 3t] =
3
,
s2 +36
s−2
,
(s−2)2 +9
19.6 L[e4t (t2 + 3t + 5)] =
19.7 L−1 [ s210
+
+25
s > 0.
4
s−3 ]
s > 2.
2
(s−4)3
+
3
(s−4)2
+
5
s−4 ,
s > 4.
= 2 sin 5t + 4e3t , t ≥ 0.
5
5 3t 3
19.8 L−1 [ (s−3)
t ≥ 0.
4 ] = 6e t ,
19.9
−1
L
19.10 L−1 [ e
e−2s
[
] = e9(t−2) h(t − 2) =
s−9
−3s (2s+7)
s2 +16
19.11 Note that
0,
e9(t−2) ,
0≤t<2
t≥2
] = 2 cos 4(t − 3)h(t − 3) + 74 sin 4(t − 3)h(t − 3), t ≥ 0.

 0, 0 ≤ t < 1
1, 1 ≤ t < 3
f (t) =

2,
t≥3
The graph of f (t) is shown below. Using Table L we find
L[f (t)] = L[h(t − 1)] + L[h(t − 3)] =
e−s e−3s
+
, s > 0.
s
s
215
19.12 Note that

 0, 0 ≤ t < 1
t, 1 ≤ t < 3
f (t) =

0,
t≥3
The graph of f (t) is shown below. Using Table L we find
L[f (t)] =L[(t − 1)h(t − 1) + h(t − 1) − (t − 3)h(t − 3) − 3h(t − 3)]
=L[(t − 1)h(t − 1)] + L[h(t − 1)] − L[(t − 3)h(t − 3)] − 3L[h(t − 3)]
=
e−s e−s e−3s 3e−3s
+
− 2 −
, s > 1.
s2
s
s
s
19.13 Note that

 0, 0 ≤ t < 1
3, 1 ≤ t < 4
f (t) =

0,
t≥4
The graph of f (t) is shown below. Using Table L we find
L[f (t)] = 3L[h(t − 1)] − 3L[h(t − 4)] =
19.14 Note that
f (t) =


3e−s 3e−4s
−
, s > 0.
s
s
0,
0≤t<1
|2 − t|, 1 ≤ t < 3

0,
t≥3
216
ANSWER KEY
The graph of f (t) is shown below. Using Table L we find
L[f (t)] =(2 − t)h(t − 1) + 2(t − 2)h(t − 2) − (t − 2)h(t − 3)
=L[−(t − 1)h(t − 1) + h(t − 1) + 2(t − 2)h(t − 2) − (t − 3)h(t − 3) − h(t − 3)]
= − L[(t − 1)h(t − 1)] + L[h(t − 1)] + 2L[(t − 2)h(t − 2)]
−L[(t − 3)h(t − 3)] − L[h(t − 3)]
=−
e−s e−s 2e−2s e−3s e−3s
+
− 2 −
+
, s > 0.
s2
s
s2
s
s
19.15 Note that
f (t) =
1, 0 ≤ t ≤ 2
0,
t>2
The graph of f (t) is shown below. From this graph we see that f (t) = h(t) − h(t −
2)h(t − 2). Using Table L we find
L[f (t)] = L[h(t)] − L[h(t − 1)h(t − 1)] =
19.16 Note that

 1, 0 ≤ t < 1
2, 1 ≤ t ≤ 4
f (t) =

1,
t>4
1 − e−2s
, s > 0.
s
217
The graph of f (t) is shown below. Using Table L we find
L[f (t)] = L[h(t − 1)] + L[h(4 − t)] =
e−s
+
s
Z
4
e−st dt =
0
1 + e−s − e−4s
, s > 0.
s
19.17 From the graph we see that
f (t) = (t − 2)h(t − 2) − (t − 3)h(t − 3) − h(t − 4)
Thus,
L[f (t)] = L[(t−2)h(t−2)]−L[(t−3)h(t−3)]−L[h(t−4)] =
e−2s − e−3s e−4s
−
, s > 0.
s2
s
19.18 From the graph we see that
f (t) = h(t − 1) + h(t − 2) − 2h(t − 3).
Thus,
L[f (t)] = L[h(t − 1)] − 2L[h(t − 3)] + L[h(t − 2)] =
e−s − 2e−3s + e−2s
, s > 0.
s
19.19 y(t) = 2e−4t +3[h(t−1)−h(t−3)]−3[e−4(t−1) h(t−1)−e−4(t−3) h(t−3)], t ≥ 0.
218
ANSWER KEY
Section 20
A2
(s−1)2
20.1 F (s) =
A1
(s−1)3
20.2 F (s) =
A1 s+A2
s2 +16
+
B1
s−2 .
20.3 F (s) =
A1 s+A2
(s2 +1)2
+
A3 s+A4
s2 +1
+
+
A3
s−1
+
+
B1
(s−2)2
B1
(s+4)2
+
+
B2
s−2 .
B2
s+4 .
A2
B1 s+B2
A1
20.4 F (s) = (s−2)
2 + s−2 + s2 +8s+17 .
h
i
1
20.5 L−1 (s+1)
= 21 e−t t2 , t ≥ 0.
3
i
h
2s−3
20.6 L−1 (s−1)(s−2)
= et + e2t , t ≥ 0.
i
h 2
+s+1
= 1 + 3 cos t + sin t, t ≥ 0.
20.7 L−1 4s
2
s(s +1)
h 2
i
20.8 L−1 s(s+6s+8
= 32 t sin 2t + 34 sin 2t − 12 t cos 2t, t ≥ 0.
2 +4)2
20.9 y(t) = 9e−2t − 6 cos 3t + 4 sin 3t, t ≥ 0.
20.10 y(t) = 4e−2t − 1 + 2t, t ≥ 0.
20.11 y(t) = 3e−2t − 2e−t + 6te−t , t ≥ 0.
20.12 y(t) = cos 2t + 12 sin 2t + 4t sin 2t, t ≥ 0.
20.13 y(t) = −et − tet + e2t , t ≥ 0.
20.14 y(t) = cos 3t + sin 3t + 23 (1 − cos 3t) − 23 (1 + cos 3t)h(t − π), t ≥ 0.
20.15 α = 1, β = 2, y0 = 2, and y00 = −3.
219
Section 21
21.1 L[f (t)] =
3−2e−s −e−2s
.
s(1−e−2s )
21.2 L[f (t)] =
2−e−s −e−3s
.
s(1−e−4s )
21.3 L[f (t)] =
e−s
[1
s2 (1−e−2s )
− (s + 1)e−s ].
21.4 L[f (t)] =
1
[1
s2 (1−e−2s )
− (2s + 1)e−2s ].
21.5 L[f (t)] =
1+e−πs
.
(1+s2 )(1−e−2πs )
21.6 L[f (t)] =
1
s
−
T = 2π.
1−e−2(s+1)
,
(s+1)(1−e−2s )
T = 2.
21.7 Note first that
e−s
1
s2 − s
+
= −
3
−s
s
s(1 − e )
s
1
se−s
−
s2 s2 (1 − e−s )
.
Using Example 21.3, we find
g(t) = 1 − f (t)
where f (t) is the sawtooth function shown below
21.8 (a) a = 2, b = 5, and c = 2. (b) F (s) = L[e−t ] =
1
s+1 ,
Y (s) = Φ(s)F (s) =
220
ANSWER KEY
t
1
.
(s+1)(2s2 +5s+2)
y(t) = −e−t + 32 e− 2 + 13 e−2t , t ≥ 0.
21.9 (a) Φ(s) =
Y (s)
F (s)
=
1
(s+1)2
(b) y(t) = L−1 [Y (s)] = 1 − e−t − te−t , t ≥ 0.
21.10 (a) Φ(s) =
Y (s)
F (s)
=
1
s2 +s+1
21.11 (a) Φ(s) =
Y (s)
F (s)
=
1
s3 −4
(b) Y (s) = Φ(s)F (s) =
(b) Y (s) =
21.12 b = 3, c = 2, y0 = 1, y00 = −1.
s2 +s−1
.
s2 (s−1)(s3 −4)
(1−e−s )
.
s(1+e−s )(s2 +s+1)
221
Section 22
22.1 (a) We have
Z
t
Z
Z
0
0
0
(b) Since F (s) = G(s) = L[h(t)] =
L−1 [ s12 ] = t, t ≥ 0.
1
s
t
ds = t, t ≥ 0.
h(t − s)h(s)ds =
f (t − s)g(s)ds =
(f ∗ g)(t) =
t
we have (f ∗ g)(t) = L−1 [F (s)G(s)] =
22.2 (a) We have
Z
t
Z
t
f (t − s)g(s)ds =
e(t−s) e−2s ds
0
0
"
#t
Z t
e(t−3s)
t
−3s
=e
e ds =
−3
0
(f ∗ g)(t) =
0
et − e−2t
, t ≥ 0.
=
3
1
1
and G(s) = L[e−2t ] = s+2
we find (f ∗ g)(t) =
(b) Since F (s) = L[et ] = s−1
1
−1
−1
L [F (s)G(s)] = L [ (s−1)(s−2) ]. Using partial fractions decomposition we find
1 1
1
1
= (
−
).
(s − 1)(s + 2)
3 s−1 s+2
Thus,
−1
(f ∗ g)(t) = L
1
[F (s)G(s)] =
3
−1
L
1
1
et − e−2t
−1
[
]−L [
] =
, t ≥ 0.
s−1
s+2
3
22.3 (a) Using the trigonometric identity 2 sin p cos q = sin (p + q) + sin (p − q) we
find that 2 sin (t − s) cos s = sin t + sin (t − 2s). Hence,
Z t
Z t
(f ∗ g)(t) =
f (t − s)g(s)ds =
sin (t − s) cos sds
0
0
Z t
Z t
1
= [ sin tds +
sin (t − 2s)ds]
2 0
0
Z
t sin t 1 t
+
sin udu
=
2
4 −t
t sin t
=
.t ≥ 0.
2
222
ANSWER KEY
(b) Since F (s) = L[sin t] =
1
s2 +1
and G(s) = L[cos t] =
(f ∗ g)(t) = L−1 [F (s)G(s)] = L−1 [
22.4 (f ∗ g)(t) =
is given below.
t2
2
(s2
s
s2 +1
we find
t
s
] = sin t, t ≥ 0.
2
+ 1)
2
2
− (t−2)
2 h(t − 2) − 2(t − 2)h(t − 2), t ≥ 0. The graph of (f ∗ g)(t)
22.5 (f ∗ g)(t) = (t − 1)h(t − 1) − 3(t − 2)h(t − 2) + 2(t − 3)h(t − 3), t ≥ 0. The
graph of (f ∗ g)(t) is given below.
22.6 t ∗ t ∗ t =
t5
120 , t
≥ 0.
22.7 h(t) ∗ e−t ∗ e−2t =
1
2
22.8 t ∗ e−t ∗ et = −t +
et
2
22.9 n = 9 and C =
− e−t + 12 e−2t , t ≥ 0.
1
8! .
22.10 y(t) = 2 + t2 , t ≥ 0.
−
e−t
2 ,t
≥ 0.
Index
Abel’s theorem, 84
Amplitude, 106
Analytic, 111
attracting solution, 62
Autonomous, 59
Bernoulli equation, 51
Characteristic equation, 91
Characteristic polynomial, 91
Complex exponential function, 103
Convolution, 168
Convolution Theorem, 168
Differential equations, 3
Direction field, 59
Equilibrium solution, 61
Euler equation, 110
Euler’s function, 103
Euler’s method, 69
Exact DE, 45
Exponential order a, 131
Exponentially bounded, 131
Extended chain rule, 43
Heun’s Method, 70
Homogeneous, 14
Initial conditions, 6
Initial value problem, 6
Integrating factor, 20
Inverse Laplace transform, 135
Laplace transform, 129
Linear combination, 81
Linear time invariant, 161
Method of integrating factor, 19
Method of Variation of Parameters, 123
Modified Euler’s method, 72
Newton’s second law, 3
Non-homogeneous, 14
Non-linear, 14
Order, 4
ordinary differential equation, 4
ordinary point, 111
Parameters, 6
Partial derivatives, 29
Partial differential equation, 5
First order linear differential equation,
Partial fractions, 151
14
Particular solution, 6
First Shifting Theorem, 143
Periodic, 159
Fundamental set of solutions, 82
Phase line, 63
Phase shift, 106
General solution, 6
Piecewise continuous, 132
Principle of superposition, 81
Heaviside step function, 141
223
224
Qualitative, 59
Recurrence formula, 114
Repelling, 62
Riccati Equation, 53
Second Shifting Theorem, 143
Semi-stable, 62
Separable, 35
Singular point, 111
Singular solution, 7, 39
Slope field, 59
Solution, 5
Solution curve, 7
Stable solution, 62
Superposition principle, 14
System transfer function, 162
Uniform motion, 3
Unstable, 62
Wave equation, 5
Wronskian, 82
INDEX