Chapter 2 - Linear Equations of Higher
Order
2.1: SecondSecond-Order Linear Equations
► d2y/dx2
= f(x,
f(x, y, dy/dx)
dy/dx)
ƒ 2 initial values: y(x0)=y0 and y’
y’(x0)=y0’
Math 215 Chapter 2
Notes and Homework
► Linear
Second Order:
A(x)y’’
A(x)y’’ + B(x)y’
B(x)y’ + C(x)y = F(x)
F(x)
► Nonlinear: Not in that form
Linear Equations of Higher Order
► Homogeneous
Linear: F(x)
F(x) = 0
► Nonhomogeneous: F(x)
F(x) ≠ 0
Differential Operators: New Notation
►D
= linear differential operator = function on
functions
► Rewrite:
y’’+p(x)y
’’+p(x)y’’ + q(x)y = 0
ƒ [D2 + p(x)D + q(x)]y = 0
► Ex:
Ex:
►
Let L = D3 – xD2 + x3.
ƒ y’’ + p(x)y’
p(x)y’ + q(x)y = f(x),
f(x), y(a)
y(a) = b0, y’(a)
(a) = b1
ƒ p, q, and f are continuous on an open interval I containing the
point a
►
Homogeneous Equations with
Constant Coefficients
► ay’’
ay’’
+ by’
by’ + cy = 0
► Look for exponential solutions: y = ert
ƒ Substitute…
Substitute…
► ar2
+ br + c = 0 is the characteristic
equation
ƒ 2 solutions: r1 and r2
ƒ Three Possibilities:
► Solutions:
Math 215 - Fall 2007
Constant Coefficient Examples
►
Solve the following:
1. Distinct Real Roots
a)
b)
+
c 2e r 2 x
y’’ + 5y’
5y’ + 6y = 0
y’’ + 5y’
5y’ + 6y = 0; y(0) = 2; y’
y’(0) = 3
2. Repeated Real Roots
a)
real; repeated real; complex conjugates
y=
Example 2:
2:
ƒ y’’ – 2y’
2y’ + y= 0
ƒ Verify that the functions y1 = ex and y2 = xex
are solutions of the differential equation .
ƒ Then find a solution satisfying the initial
conditions y(0) = 3, y’
y’(0) = 1.
of Superposition (Theorem 1)
c1er1x
THEN
► 2.1
ƒ L(c1y1 + c2y2) = c1Ly1 + c2Ly2
ƒ Hence if y1 and y2 are two solutions to the linear
differential equation, then c1y1 + c2y2 is also.
►Distinct
IF
ƒ there is a unique solution on the interval I
ƒ Find L(sin 2x) and L(e-4x)
► Principle
Existence and Uniqueness (Thm
(Thm 2)
y’’ + 6y’
6y’ + 9y = 0
3. Complex Roots
a)
9y’’
9y’’ + y = 0
1
Chapter 2 - Linear Equations of Higher
Order
Linear Independence
Two functions f1 and f2 are linearly dependent
on an interval I if there exist constants c1 and c2,
with at least one ci ≠ 0, such that for all x in I
c1f1(x) + c2f2(x) = 0.
F and g are linearly independent if they are
not linearly dependent.
►
►
ƒ
►
The Wronskian
►
W ( y1, y 2 ) =
Ex:
Ex: Determine whether the following pairs of
functions are linearly dependent or independent:
The Wronskian and L.I.
W=
y 1( x ) y 2 ( x )
y'1 ( x ) y'2 ( x )
►
Thm:
Thm: If f and g are differentiable functions on I
and W(f,g)(x0)≠0 for some x0 ∈ I, then f and g
are linearly independent on I.
►
Ex:
Ex: Are et and e2t linearly independent?
Ex:
Ex: Show t and t5 are linearly independent on
-∞ < t < ∞
►
►
y 1( x ) y 2 ( x )
y'1 ( x ) y'2 ( x )
Ie:
Ie: if the only constants that work are c1 = c2 = 0.
►
NOTE:
NOTE:
It is possible that W = 0 and the functions are still
linearly independent: t3 and |t3| on -∞ < t < ∞
Ex:
Ex: Find the wronskian for the following
pairs of functions
1. sin x and cos(x – π/2)
2. ex and e2x
1. sin x and cos(x – π/2)
2. ex and e2x
►
The Wronskian of two functions y1 and
y2 is given by
Existence of Independent Solutions
►
Suppose that y1 and y2 are solutions of y’’
y’’
+ p(x)
p(x) y’
y’ + q(x)
q(x) y = 0 on an open interval
I on which p and q are continuous.
ƒ
Thm 3 (Wronskian of Solutions)
1.
2.
ƒ
If y1 and y2 are linearly dependent, then W(y1, y2)
≡ 0 on I.
If y1 and y2 are linearly independent, then W(y1, y2)
≠ 0 at each point of I.
Thm 4 (General Solutions)
►
If Y is any solution on I, then there exist numbers
c1 and c2 such that Y(x)
Y(x) = c1 y1(x) + c2 y2(x).
2.2: General Solutions
of Linear Equations
Change of Variables: Euler Equation
► Consider
ax2y’’ + bxy’
bxy’ + cy = 0, x > 0,
where a, b, and c are real constants
► (HW
#51) Consider the substitution v = ln t
ƒ Show: dy/dx = 1/x dy/dv
ƒ Show: d2y/dx2 = 1/x2 [d2y/dv2 – dy/dv]
dy/dv]
ƒ Show the equation becomes:
ad2y/dv2 + (b – a)dy/dv + cy = 0
► Ex:
Ex:
Solve x2y’’ – 4xy’
4xy’ – 6y = 0.
Math 215 - Fall 2007
► General
Nonhomogeneous Form:
ƒ Ly = f(x)
f(x)
ƒ [Dn + p1(x)Dn-1 + … + pn-1(x)D + pn(x)]y = f(x)
f(x)
► Required
Initial Condition:
(n-1)
(n-1)(x ) = y (nƒ y(x0) = y0; y’
y’(x0) = y0’; …; y(n0
0
ƒ ie:
ie: Must have first nn-1 derivatives specified
► Existence
and Uniqueness (Thm
(Thm 2)
ƒ If p1, p2, …, pn, and f are continuous on I, then there
exists exactly one solution of Ly = f(x)
f(x) that also satisfies
the initial condition.
2
Chapter 2 - Linear Equations of Higher
Order
Nth Order Wronskian
y1
y1'
W ( y1,..., y n ) =
M
y (1n−1)
y 2 L yn
y 2 ' L yn '
M
M
y (2n−1) L y (nn−1)
► If
the coefficients of Ly = 0 are continuous on I
and y1, …, yn are solutions of Ly = 0, and if
W(y1, …, yn)(t)
)(t) ≠ 0 for at least one point in I, then
every solution of Ly = 0 can be expressed as a
linear combination of y1, …, yn.
ƒ Ie:
Ie: y1, …, yn form a fundamental solution set
ƒ Ex:
Ex: Determine the Wronskian for 1, x, ex
Linear Independence
► y1,
…, yn form a fundamental set
ƒ Linear combination c1y1 + … + cnyn form the
general solution.
solution.
► y1,
…, yn are linearly dependent if there exist
constants c1, …, cn, not all zero, such that c1y1 +
… + cnyn = 0
► Otherwise, they are linearly independent
► Ex:
Ex: Determine whether 2t – 3, 2t2 + 1, and 3t2
t are linearly dependent or independent.
► NOTE:
If y1, …, yn are solutions to Ly = 0, then
they are linearly independent if W(y1, …, yn) ≠ 0.
Reduction of Order
Nonhomogeneous Equations
► Nonhomogeneous:
► Associated
► General
Ly = f(x)
f(x)
Homogeneous: Ly = 0
Solution to Nonhomogeneous
ƒ y = yc + yp
= (solution to homogeneous) + (particular solution)
► 2.2
Example 7:
7: y’’
y’’ + 4y = 12x.
ƒ yp = 3x is a particular solution
ƒ Find yc and the general solution y(x)
y(x)
ƒ Find a solution that satisfies y(0) = 5, y’
y’(0) = 7
2.3: Homogeneous Equations with
Constant Coefficients
► The
Constant Coefficient Linear Equation
ƒ [anDn + a1Dn-1 + … + a0]y = 0
► The
Characteristic Equation
ƒ anrn + a1r n-1 + … + a0 = 0
ƒ Factoring: Suppose anrn + a1r n-1 + … + a0 = 0 has
integer coefficients. Then if r = p/q,
p/q, p is a factor of a0
and q is a factor of an
► Types
of Roots from Factoring Characteristic
Polynomial
ƒ Real, Repeated,
Repeated, or Complex
Math 215 - Fall 2007
+
► Let y1(x)
be one solution of
y’’ + p(x)y’
p(x)y’ + q(x)y = 0
ƒ Find another solution: let y = v(x)y1(x)
ƒ Substitution yields 1st order equation for v’(x)
(x)
►Solve
for v(x)
v(x)
► Ex:
Ex:
Given y1(x) = x is a solution of
(x2D2 + xD – 1)y = 0,
find a second solution.
► Ex:
Ex: Given y1(x) = x-1 is a solution of
2x2y’’ + 3xy’
3xy’ – y = 0, x > 0,
find a second linearly independent solution.
Repeated Roots
► Solve
y’’
y’’ + 4y’
4y’ + 4y = 0
ƒ y1 = e-2x
ƒ Hence cy1 is a solution. Look for solutions of
form c(t)y1
ƒ Check solutions for linear independence
► Solutions
to equations with repeated root r
will have form y = c1erx + c2xerx
► Ex:
Ex: Solve y’’
y’’ – 2y’
2y’ + y = 0, y(0) = 2, y’
y’(0) = 1
3
Chapter 2 - Linear Equations of Higher
Order
Euler’s Formula
Nth Order ODE w/ Constant Coef’s
►
e(a + bi) = ea(cos b + i sin b)
►
Suppose the characteristic equation of ay’’
ay’’ + by’
by’
+ cy = 0 has roots r = a ± bi
ƒ
ƒ
ƒ
►
► Types
Write solution in exponential form
Rewrite using Euler’
Euler’s formula
Check for linear independence of parts
y(0) = -2;
ƒ Real: r → ert
ƒ Complex: a ± bi → eat cos bt and eat sin bt
ƒ Repeated: Add terms with increased order of t
► Examples:
Examples:
Ex:
Ex: Find the general solution of
1. y’’ + y’
y’ + y = 0
2. y’’ + 9y = 0
3. 16y’’
16y’’ – 8y’
8y’ + 145y = 0;
of Roots from Factoring Characteristic
Polynomial
y’(0) = 1
2.4: Mechanical Vibrations
► Tools
Basic Model: The Spring
for Spring Models
ƒ Newton: Ftotal = ma = mx’’
(t))
mx’’(t
ƒ Hooke’
Hooke’s Law: Spring force is proportional to elongation.
Fspring = -k ⋅ (displacement) = -kx
ƒ Weight: Fweight = mg
ƒ Damping/Resistive Force: Resistive force is proportional
to velocity. Fdamp = -c x’(t)
(t)
ƒ External force: Fext(t)
(t)
► Model:
Ftotal = Fspring + Fweight + Fdamp + Fext
Solve the following ODEs
ƒ y(4) + y’’’
y’’’ – 7y’’
7y’’ – y’ + 6y = 0
ƒ y(4) – y = 0
ƒ y(4) + 2y’’
2y’’ + y = 0
► Ex:
Ex:
A mass of 4 kg stretches a spring 2 cm.
If the mass is displaced an additional 6 cm
in the positive direction and then released,
and if there is no damping, set up the
differential equation to determine the
position of the mass at any time t.
ƒ ODE?
ƒ Initial conditions?
ƒ If there is no external force, is this equation
homogeneous?
Example 1: Undamped Spring
Undamped Spring
► Solve:
(mD2 + k)x = 0
cos(a
cos(a – b) = cos a cos b + sin a sin b
rewrite solution in form x = C cos(ω
cos(ω0t – α)
► Using
ƒ
ƒ
ƒ
ƒ
ƒ
Period:
Period: T = 2π
2π/ω0
Circular Frequency of vibration: ω0
Amplitude of motion: C
Phase Angle:
Angle: α
Time lag:
lag: δ = α/ω0
► Observations
ƒ
ƒ
ƒ
ƒ
Amplitude does not diminish as t → ∞
Initial conditions determine amplitude
T increases as m increases
T decreases as k increases
Math 215 - Fall 2007
►A
body with mass m = ½ kg is attached to the
end of a spring that is stretched 2m by a force of
100 N. It is set in motion with initial position x0 =
1 m and initial velcity v0 = -5 m/s.
m/s.
► Determine
the
position function
of the body as
well as the
amplitude,
frequency,
period, and time
lag of its motion.
5
0
0.5
1
1.5
2
4
Chapter 2 - Linear Equations of Higher
Order
2.5: Nonhomogeneous Equations
and Undetermined Coefficients
Damped Spring
► Find
the roots of the characteristic equation
of mx’’
mx’’ + cx’
cx’ + kx = 0.
ƒ Alternatively: x’’
x’’ + 2px’
2px’ + ω02x = 0
ƒ ω0 = (k/m)1/2 = undamped circular frequency
ƒ p = c/(2m) > 0
►3
Cases For Discriminant:
Discriminant:
ƒ>0
ƒ=0
ƒ<0
Motion is overdamped
Motion is critically damped
Motion is underdamped
►
Nonhomogeneous Equation
►
Thm:
Thm: If Y1 and Y2 are solutions of the nonhomogeneous
equation, then Y1 – Y2 is a solution of the corresponding
homogeneous.
►
Thm:
Thm: If y1 and y2 are a fundamental set of solutions of the
homogeneous, then Y1(t) – Y2(t) = c1y1(t) + c2y2(t) where
c1 and c2 are constants.
►
Thm:
Thm: Solution to the nonhomgeneous can be written in
form y = c1y1(t) + c2y2(t) + Y(t)
Y(t) where y1, y2 are a
fundamental set of the homogeneous and Y is specific to
the nonhomogeneous (called the particular soln).
►Damping
►x
is small
= Ce-ptcos(ω
cos(ω1t – α) where ω1 = (4km – c2)1/2/(2m)
Method of Undetermined Coefficients
PrePre-Ex:
Ex: Solve (D – 2)(D2 – 3D – 4)y = 0.
►
Summary: Undetermined Coeffs
► To
Solve Nonhomogeneous Linear with Constant
Coefficients
Ex 1:
1: Find the solution of y’’
y’’ – 3y’
3y’ – 4y = 3e2t.
►
ƒ Ly = f(x)
f(x)
ƒ f(x)
f(x)
► Constant
► Exponential: erx
► Trig: sin kx,
kx, cos
Steps:
ƒ
1.
2.
Solve the related homogeneous Ly = 0
Find an annihilator A(D) for the right side:
► A(D)g(t)
A(D)g(t) = 0
3.
4.
5.
6.
7.
►
Determine form of Y(t)
Y(t) from general solution of A(D)Ly = 0
Drop terms already satisfying Ly = 0
Substitute simpler guess into Ly = g(t)
g(t)
Determine coefficients that yield g(t)
g(t) on right
Solution is from (1) and (6)
ƒ Steps:
1. Solve the related homogeneous Ly = 0
2. Find an annihilator A(D) for the right side: A(D)f(x)
A(D)f(x) = 0
3. Determine form of Y(x)
Y(x) from general solution of A(D)Ly =
4. Drop terms already satisfying Ly = 0
5. Substitute simpler guess into Ly = f(x)
f(x)
6. Determine coefficients that yield f(x)
f(x) on
Ex 2:
2: Find the solution of y’’
y’’ – 2y’
2y’ – 3y = 6 – 8et
7. Solution
Find the annihilators A(D) for each of the
following that would allow you to find the
general solution
1.
2.
3.
4.
5.
y’’ – 3y’
3y’ – 4y = 2sin x
y’’ – 3y’
3y’ – 4y = -8ex cos 2x.
y’’ – 3y’
3y’ – 4y = 2 sin x – 8ex cos 2x
y’’ – 3y’
3y’ – 4y = 2e-x
y’’ – 4y = ex + 2e2x
Math 215 - Fall 2007
is from (1) and (6)
0
right
Undetermined Coefficients
More Examples
►
kx
► Exponentials times trig: erxcos kx,
kx, erxsin kx
► Positive integer powers of x times any of these
Recall:
►
Find solutions to Homogeneous, yh
Find A(D)
Apply A(D) to find form of particular solution, Y
Substitute Y into original equation to find coefficients
Write general solution = yh + Y
ƒ
ƒ
ƒ
ƒ
ƒ
►
Examples
1.
2.
3.
y’’’ – 3y’’
3y’’ + 3y’
3y’ – y = 4ex
y(4) + 2y’’
’’
+
y = 3 sin x – 5 cos x
2y
y’’’ – 4y’
4y’ = x + 3 cos x + e-2x
5
Chapter 2 - Linear Equations of Higher
Order
Variation of Parameters
►
Steps: Variation of Parameters
The good news
►
2.
The bad news
ƒ
ƒ
ƒ Leads to n equations in n unknowns (for nth order ODE)
ƒ Can lead to difficult integrals.
► Consider
Solve homogeneous
Solve the system for u’
u’1 and u’
u’2:
1.
ƒ Requires only that we know the general solution to Ly = 0
Integrate to find u1 and u2
Particular solution: Y(x)
Y(x) = u1(x)y1(x) + u2(x)y2(x)
General solution = homogeneous solution +
particular solution
3.
[D2 + p(x)D + q(x)]y = f(x)
f(x)
4.
ƒ Solve homogeneous: y = c1y1(x) + c2y2(x)
ƒ Let coefficients vary: y = u1(x)y1(x) + u2(x)y2(x)
ƒ Substitute into nonhomogeneous
5.
u’
u’1(x)y1(x) + u’
u’2(x)y2(x) = 0
yields: u’
u’1(x)y’
(x)y’1(x) + u’
u’2(x)y’
(x)y’2(x) = f(x)
f(x)
► Why do we know this system must have a solution?
► Let
► Substitution
Ex 1:
1: Solve 4y’’
4y’’ – 4y’
4y’ + y = x1/2ex/2, x > ∞
Ex 2:
2: Solve (D2 + 1)y = sec x, -π/2 < x < π/2
►
ƒ Solve the system for u’
u’1 and u’
u’2.
ƒ Integrate to find u1 and u2
u’1(x)y1(x) + u’
u’2(x)y2(x) = 0
u’1(x)y’
(x)y’1(x) + u’
u’2(x)y’
(x)y’2(x) = f(x)
f(x)
►
Steps: Variation of Parameters
1.
2.
2.7: Electrical Circuits
= current (amperes)
► E(t)
E(t) = voltage (volts)
► Q = charge (coulombs)
► Elementary Laws of Electricity
a)
Solve the system for u’
u’1 , …, u’n:
►
►
►
►
b)
c)
u’1y1 + u’
u’2y2 + … + u’nyn = 0
u’1y’1 + u’
u’2y’2 + … + u’ny’n = 0
…
(n-1) + u’
(n-1) + … + u’
(n-1) = g
u’1y(nu’2y(nu’ny(n1
2
n
General solution = homogeneous + particular
►
Ex:
Ex: Use variation of parameters to solve:
y’’’ + y’
y’ = sec x
R
Circuit Example
► Kirchoff’
Kirchoff’s
C
I
L
E(t)
Law
Homework
L
a formula for the charge Q at time t
seconds on the capacitor of a simple circuit
with no external voltage (E(t
(E(t)) = 0),
resistance R = 2 ohms, capacitance C = 1
farad, and inductance = 1 henry,
henry, given that
charge Q = 0 and current I = 1 ampere at
time t = 0.
Math 215 - Fall 2007
I
ƒ The sum of the voltage drops in a simple loop of an
electrical circuit is equal to the applied voltage.
E(t)
► Find
C
ƒ Voltage drop across a resistor is IR
R (measured in ohms)
ƒ Voltage drop across an inductor is L dI/dt
L (measured in henrys)
ƒ Voltage drop across a capacitor is Q/C
C (measured in farads)
Integrate to find u1 , …, un
Write Particular solution: Y(x)
Y(x) = u1(x)y1(x) + … + un(x)yn(x)
(x)
3.
R
►I
Solve homogeneous: Fundemental Set = y1, …, yn
Find Particular Solution: Y = u1(x)y1(x) + … + un(x)yn(x)
(x)
► 2.1:
#3, 7, 9, 15, 17, 20, 22, 27, 28, 31, 33,
39, 43, 45, 51, 52
► 2.2: #1, 7, 9, 13, 21, 25, 33, 38, 41
► 2.3: #1, 5, 6, 9, 11, 16, 21, 29, 35, 39, 41,
52
► 2.4: #1, 3, 13, 14, 15, 17, 19, 22, 23
► 2.5: #1, 5, 7, 15, 17, 25, 26, 27, 28, 29, 31,
4747-55 odd
► 2.7: #1#1-9 odd, 17, 19
6