Second Order Equations
Today, we will begin a discussion of solving second order linear equations. We will
discuss some of the theory of second order linear homogeneous equations. We will use
existence and uniqueness theorem to show that we can write down all possible solutions to these equations in terms of a set of so-called fundamental solutions. Along
the way, we will uncover a tool (the Wronskian) for determining when we have a
fundamental set of solutions.
We then write the general solution to a second-order linear homogeneous equation
if we can find a fundamental set of solutions. We will begin solving a second order homogeneous equation with constant coefficients by finding fundamental sets of
solutions.
1. Introduction
2. Existence and Uniqueness
3. General Solutions for Second Order Equations
4. The Wronskian and Fundamental Solutions
5. Solving Constant Coefficient Equations
1
Introduction
We will consider second order linear equations of the form
y ′′ + p(t)y ′ + q(t)y = g(t)
where p, q, and g are arbitrary functions of the independent variable t. We can of
course convert equations of the form
P (t)y ′′ + Q(t)y ′ + R(t)y = G(t)
into this form by dividing by P (t), whenever P (t) 6= 0. Points where P (t) = 0 are
called singular points, and we will not be discussing methods of dealing with these
points.
1
We will be working mostly with equations which have constant coefficients (i.e., where
p(t) and q(t) are constant). We will first be considering homogeneous equations
(those for which G(t) = 0), and then later describing how to solve non-homogeneous
equations.
2
Existence and Uniqueness
We have an existence and uniqueness theorem for second order linear equations which
is quite similar to the theorem for first order equations:
Theorem 3.1: (p. 111) Let p(t), q(t), and g(t) be continuous
functions on the interval (a, b) and let t0 be in (a, b). Then the initial
value problem
y ′′ + p(t)y ′ + q(t)y = g(t),
y ′ (t0 ) = y0′
y(t0 ) = y0 ,
has a unique solution which is defined on the entire interval (a, b).
We will not attempt to prove this theorem, but we will use it later.
Example: Find the largest interval in which the initial value problem
(t − 1)y ′′ − 3ty ′ + 4y = sin(t),
y(−2) = 2,
y ′ (−2) = 1
is certain to have a unique solution.
In standard form, we have
y ′′ −
4
sin(t)
3t ′
y +
y=
t−1
t−1
t−1
for which the coefficient functions are continuous when t 6= 1. Thus, given that the
initial value is at t0 = −2, we have a unique solution at least on (−∞, 1).
There is an equivalent theorem (Theorem 4.1, p. 224) which applies to a general nth
order linear differential equation.
3
General Solutions
We are interested in finding the general solution to a homogeneous second order linear
differential equation. Recall that a second order equation should allow us to set two
initial conditions, so an initial value problem looks like the following:
y ′′ + p(t)y ′ + q(t)y = 0,
y(t0 ) = y0 ,
y ′(t0 ) = y0′ .
What might a general solution to the differential equation look like? We might
expect that since there are two initial conditions we can satisfy, a general solution
might involve two unknown constants we can set. In fact, this is the case.
Better than that, we will see soon that there will be two different functions which we
can combine to create a solution to a second order linear equation. The technique
will depend on the principle of superposition (see Theorem 3.2, p. 116):
If y1 (t) and y2 (t) are both solutions to the homogeneous second order
equation
y ′′ + p(t)y ′(t) + q(t)y(t) = 0
(1)
then any linear combination of y1 and y2 (c1 y1 (t) + c2 y2 (t) where c1 and
c2 are constants) is also a solution to y ′′ + p(t)y ′ (t) + q(t)y(t) = 0.
The principle of superposition is easy to prove: just assume that y1 and y2 are solutions
to equation (4) above, and let y(t) = c1 y1 (t) + c2 y2 (t). Then we see that
d
d2
[c
y
+
c
y
]
+
p
[c1 y1 + c2 y2 ] + q [c1 y1 + c2 y2 ]
1
1
2
2
dt2
dt
= c1 y1′′ + c2 y2′′ + p c1 y1′ + p c2 y2′ + q c1 y1 + q c2 y2
= c1 [y1′′ + py1′ + qy1 ] + c2 [y2′′ + py2′ + qy2 ] = 0
y ′′ + py ′ + qy =
since both y1 and y2 are solutions to (4). The principle of superposition is wonderfully
useful, and we will refer to it often. (Note that it does not work for non-homogeneous
equations!)
For the moment, it suggests that if we have two different solutions to a second order
linear homogeneous equation, we might be able to write a general solution as c1 y1 (t)+
c2 y2 (t). (After all, this function does have two unknown constants.)
For example, let’s consider the following initial value problem:
y ′′ + 8y ′ − 9y = 0,
y(0) = 1,
y ′(0) = −19.
It is easy to show that both et and e−9t are solutions to this differential equation:
d2 t
e
dt2
+ 8 dtd et − 9et = et + 8et − 9et = 0
d2 −9t
e
dt2
+ 8 dtd e−9t − 9e−9t = (−9)e−9t + 8(−9)e−9t − 9e−9t = 0
We did not say how we came up with these two solutions. We will consider that
question later. We suggest that the general solution might be
y(t) = c1 y1 (t) + c2 y2 (t) = c1 et + c2 e−9t ,
meaning that we think every solution might be of this form. Now this has two
constants available, so perhaps we can solve for c1 and c2 from the initial conditions.
We see that we need
y(0) = c1 + c2 = 1
and as y ′ (t) = c1 et − 9c2 e−9t ,
y ′ (0) = c1 e0 − 9c2 e0 = c1 − 9c2 = −19
Thus, we have two linear equations with two unknowns:
c1 + c2 = 1
c1 − 9c2 = −19
This is easy to solve. If we solve the first equation for c1 , we get c1 = 1 − c2 . Plugging
this into the second equation gives
(1 − c2 ) − 9c2 = −19,
or
− 10c2 = −20
so c2 = 2. Then c1 = 1 − c2 = −1. So the particular solution to the IVP is
y(t) = −et + 2e−9t .
We have successfully solved a second order, homogeneous, linear initial value problem by using two different solutions to the differential equation. Next time, we will
consider whether such an approach will always work or not.
4
General Solutions for Second Order Equations
Last time, we suggested that we might be able to solve a second-order, linear, homogeneous initial value problem by finding two solutions to the differential equation
y1 (t) and y2 (t), and taking some linear combination of these two. In other words, we
suggested that the general solution to a second-order, linear, homogeneous equation
might be of the form
y(t) = c1 y1 (t) + c2 y2 (t)
where y1 (t) and y2 (t) are two solutions to the differential equation. As an example,
we showed that both y = et and y = e−9t were solutions to the differential equation
y ′′ + 8y ′ − 9y = 0, and showed that we could solve the initial value problem
y ′′ + 8y ′ − 9y = 0,
y(0) = 1,
y ′(0) = −19.
with the linear combination
y(t) = −et + 2e−9t .
Will this approach always work? What if we tried et and 2et as our two solutions
to y ′′ + 8y ′ − 9y = 0? (The principle of superposition after all assures that 2et is a
solution if et is.) Then our “general solution” would look like
y = c1 et + c2 2et = (c1 + 2c2 )et .
If we try to solve the same two initial conditions as before, we see that y(0) = 1 gives
the equation c1 + 2c2 = 1, and y ′(0) = −19 gives c1 + 2c2 = −19. But there is no
solution to this system of equations!
The problem of course is that et and 2et are not really “different” functions; one
is just a constant multiple of the other. How will we decide which functions are
“really different”, and can be used to form a general solution?
5
The Wronskian and Fundamental Solutions
Suppose we have a second order differential equation
y ′′ + p(t)y ′ + q(t)y = g(t)
(2)
and a set of two solutions, y1 (t) and y2 (t). According to the discussion above, we
would like to know if the two solutions are “really different”, and if we can express
all solutions to equation (6) in the form
y(t) = c1 y1 (t) + c2 y2 (t).
If the above truly forms the general solution to equation (6), we will say y1 and
y2 form a fundamental set of solutions for equation (6). In other words, y1 and y2
form a fundamental set of solutions for equation (6) if every solution is of the form
c1 y1 (t) + c2 y2 (t).
Suppose that y(t) = c1 y1 (t) + c2 y2 (t) is a general solution. (In other words, we can
solve every initial value problem involving this differential equation using this function
y.) Let’s try to solve the initial value problem
y(t) = c1 y1 (t) + c2 y2 (t),
y ′ (t0 ) = y0′ .
y(t0 ) = y0 ,
In other words, we want to find c1 and c2 such that
c1 y1 (t0 ) + c2 y2 (t0 ) = y0
c1 y1′ (t0 ) + c2 y2′ (t0 ) = y0′
We put this in the form of a matrix equation below:
"
y1 (t0 ) y2 (t0 )
y1′ (t0 ) y2′ (t0 )
# "
·
c1
c2
#
=
"
y0
y0′
#
Then we can solve for any right hand side exactly when the determinant
y1 (t0 ) y2 (t0 )
y1′ (t0 ) y2′ (t0 )
(3)
does not equal zero. Call the determinant (3) above the Wronskian of y1 and y2 at
the point t0 , denoted W (t0 ) or sometimes W (y1 , y2)(t0 ).
Thus, we can solve a linear second order homogeneous initial value problem (with
initial conditions set at t = t0 ) if we have two solutions y1 and y2 whose Wronskian
W (y1 , y2)(t0 ) is non-zero.
We claim that every solution to the differential equation has the form y(t) = c1 y1 (t) +
c2 y2 (t) for some constants c1 and c2 . Then the set of functions y1 and y2 are a
fundamental set of solutions (since every solution can be written in terms of these),
and y(t) = c1 y1 (t) + c2 y2 (t) is the general solution.
Theorem:
Let y1 (t) and y2 (t) be solutions of the homogeneous
linear differential equation
y ′′ + p(t)y ′ + q(t)y = 0,
a < t < b,
where p(t) and q(t) are continuous on (a, b). Let W (t) denote the Wronskian of y1 and y2 . If there is a point t0 in (a, b) where W (t0 ) 6= 0, then
y1 and y2 form a fundamental set of solutions.
Proof: Suppose we have a solution φ(t) to our differential equation y ′′ + p(t)y ′ +
q(t)y = 0 defined about some point t1 contained in interval (a, b) where our coefficient
functions p and q are continuous. We know from the existence theorem that there is
a unique solution to the initial value problem
y ′′ + p(t)y ′ + q(t)y = 0,
y(t1 ) = φ(t1 ),
y ′(t1 ) = φ′ (t1 ).
Since φ is obviously such a solution, it must be the solution guaranteed by the theorem, and therefore φ(t) is defined on all of (a, b).
If y1 and y2 are two solutions to y ′′ + p(t)y ′ + q(t)y = 0 for which W (y1, y2 )(t0 ) 6= 0
for some point t0 in (a, b), we will now show that φ must be of the form c1 y1 (t)+c2 y2 (t).
Since φ must exist on all of (a, b), it must exist at t = t0 . Let y0 = φ(t0 ) and
y0′ = φ′ (t0 ). The fact that W (y1 , y2 )(t0 ) 6= 0 assures us that we may solve
c1 y1 (t0 ) + c2 y2 (t0 ) = y0
c1 y1′ (t0 ) + c2 y2′ (t0 ) = y0′
for constants c1 and c2 . Therefore, we can solve the initial value problem
y ′′ + p(t)y ′ + q(t)y = 0,
y(t0 ) = y0 ,
y ′ (t0 ) = y0′
using a linear combination y(t) = c1 y1 (t) + c2 y2 (t). But then both y(t) and φ(t) are
solutions to this initial value problem—and by uniqueness, y(t) = φ(t)!
So we now know that all we need to do is find a set of fundamental solutions, and
we know we can determine if we have found fundamental solutions by checking the
Wronskian.
Example:
Suppose we have y1 (x) = e2x and y2 = xe2x . We can confirm that both are solutions
to
y ′′ − 4y ′ + 4y = 0
(We have not yet discussed how we might have found such solutions.)
Now determine if y1 and y2 form a fundamental set of solutions. (Are they “really
different”?)
We check the Wronskian:
e2x
xe2x
2x
2x
2e
e + 2xe2x
= (2x + 1)e4x − 2xe4x = e4x 6= 0
Note that we needed to check that W (t0 ) 6= 0 for some t0 where the solution exists
(in this case, any −∞ < t0 < ∞), but in fact the Wronskian is never zero at any
point. So these form a fundamental set of solutions. Therefore, we have the general
solution
y(t) = c1 e2x + c2 xe2x
to the differential equation above. We emphasize again that any solution to y ′′ −4y ′ +
4y = 0 must be of this form.
Finally, we mention one other interesting result regarding fundamental sets and the
Wronskian:
Given two solutions to y ′′ + p(t)y ′ + q(t)y = 0 on (a, b) (p and q
continuous on (a, b) of course), then the Wronskian of these solutions
is either 0 on the entire interval (a, b), or it is non-zero on the entire
interval (a, b). (Note: This is only true if the functions are solutions to
the differential equation.)
So the result in example 5 above is not unusual. If we have a fundamental set of
solutions, the Wronskian is non-zero on the entire interval where we are guaranteed
the existence of unique solutions. (See p. 190, ”Abel’s Theorem.”)
Theorem: Suppose y1 (t) and y2 (t) are both solutions to to the homogeneous differential equation
y ′′ + p(t)y ′ + q(t)y = 0,
a<t<b
where p and q are continuous on (a, b). Then:
1. If y1 and y2 are a fundamental set of solutions, then they are linearly
independent on (a, b).
2. If y1 and y2 are not a fundamental set, they are linearly dependent
on (a, b).
Since the Wronskian will tell us whether or not two solutions form a fundamental set,
it will also tell us if the solutions are linearly independent. Note that this only applies
to functions which are solutions to a homogeneous linear differential equation; it is
possible for the Wronskian of two functions to be zero and for the functions to still
be linearly independent anyway.
So what we determined earlier is that we can write the general solution to a secondorder, homogeneous linear differential equation if we can find two linearly independent
solutions. Or equivalently, two linearly independent solutions to a second-order, homogeneous linear differential equation form a fundamental set of solutions for that
equation.
Example:
Given that e2x and 2e2x are both solutions to y ′′ − 4y ′ + 4y = 0, determine if these
two are linearly independent.
From the Wronskian, we have
e2x 2e2x
2e2x 4e2x
= 4e4x − 4e4x = 0
Since these are solutions to the differential equation, we may conclude that they are
linearly dependent, and thus do not form a fundamental set of solutions. (It should
be obvious in this case that these are linearly dependent; note that 2(e2x )−(2e2x ) = 0.)
Also note that we only need to check the values of the solutions and their derivatives at a single point to determine if they form a fundamental set (and therefore are
linearly independent), since the Wronskian of solutions is either always zero or always
non-zero on the interval of the solution.
Example:
We can show that y1 (t) = cos(2t) and y1 (t) = cos(2t − π/2) are both solutions to
y ′′ + 4y = 0. Are these solutions linearly independent? (Do they form a fundamental
set of solutions?)
The Wronskian gives
cos(2t)
cos(2t − π/2)
−2 sin(2t) −2 sin(2t − π/2)
= −2 cos(2t) sin(2t − π/2) + 2 cos(2t − π/2) sin(2t)
This is somewhat sticky to work with. However, if we look at at particular point,
such as t = 0, all we have is
y1 (0) y2 (0)
y1′ (0) y2′ (0)
cos(0)
cos(−π/2)
=
−2 sin(0) −2 sin(−π/2)
1 0 =2
=
0 2 So these do form a fundamental set of solutions, and are therefore linearly independent. We know that we can write the general solution to a second-order linear
homogeneous equation if we can find a fundamental set of solutions. Must such a
solution exist? (How many fundamental sets are there for a differential equation?)
Once we have established all second-order linear homogeneous equations have a fundamental set of solutions, we will begin solving a second order homogeneous equation
with constant coefficients by finding fundamental sets of solutions.
6
Constant Coefficients: The Characteristic Equation
We turn to the specialized case of a second order, linear, homogeneous, ordinary
differential equation with constant coefficients. (Whew!) In other words:
ay ′′ + by ′ + cy = 0
(4)
We wish to search for possible solutions to this differential equation. We hope there
might be a function which would not change much upon differentiation, so that the
function and its first two derivatives would mostly cancel out. We might then be
inspired by previous experience to try solutions of the form ert , since:
y = ert
dy/dt = rert
d2 y/dt2 = r 2 ert
We look for which values of r will make ert a solution to ay ′′ + by ′ + cy = 0, and get
the following:
0 = ay ′′ + by ′ + cy
h i
d2 h rt i
d h rt i
= a 2 e +b
e + c ert
h i
i
h dt i
hdt
2 rt
= a r e + b rert + c ert
= ert ar 2 + br + c .
Thus, if ert is a solution to equation 4, we must have
ar 2 + br + c = 0
(5)
Equation 5 is referred to as the characteristic equation of equation 4. The roots of
the characteristic equation can be obtained from the quadratic formula:
√
−b ± b2 − 4ac
r1,2 =
2a
Three cases are possible:
a)
two real roots when
b2 − 4ac > 0
b) one repeated root when b2 − 4ac = 0
c) two complex roots when b2 − 4ac < 0
If the characteristic equation gives us two real roots r1 and r2 , then we get two
solutions to equation 4:
y1 (t) = er1 t ,
and y2 (t) = er2 t .
Then the principle of superposition tells us that any linear combination of y1 and y2
will be a solution to equation 4. We then should have two different solutions to the
characteristic equation, r1 and r2 . (So r1 6= r2 .) Two solutions to the differential
equation then are
y1 (t) = er1 t and
y2 (t) = er2 t .
Do these form a fundamental set?
We check the Wronskian:
er1 t
er2 t
r1 er1 t r2 er2 t
= r2 er1 t0 er2 t0 − r1 er2 t0 er1 t0 = e(r1 +r2 )t0 (r2 − r1 ) 6= 0.
The last is true because we assumed r1 6= r2 . So as long as we have two different, real
roots to the equation, we now know how to find the general solution to the equation.
(This also means of course that we can solve initial value problems of this type.)
Example:
Find two solutions to the differential equation
y ′′ + 8y ′ − 9y = 0.
We start by assuming solutions of the form y = ert . Such solutions must satisfy the
characteristic equation, which in this case is
r 2 + 8r − 9 = 0
The characteristic equation has solutions
r1 = 1
and
r2 = −9
Therefore, two solutions to the equation are
y1 (t) = et
and
y2 (t) = e−9t
Finally, any linear combination is a solution, so we know
y(t) = c1 et + c2 e−9t
is also a solution.
Here’s another example:
Example:
Find the solution to the initial value problem
4y ′′ − y = 0,
y(−2) = 1,
y ′(−2) = −1.
First we find the characteristic equation:
4r 2 − 1 = 0
The characteristic equation has solutions
r1 = 1/2
and
so we have the two solutions
r2 = −1/2
y1 (t) = er1 t = et/2 and y2 (t) = er2 t = e−t/2
and thus a general solution
y(t) = c1 er1 t + c2 er2 t = c1 et/2 + c2 e−t/2
To solve the initial value problem, we need
y(−2) = 1 = c1 e1 + c2 e−1
and
y ′(−2) = −1 = c1 (1/2)e1 + c2 (−1/2)e−1
These equations have solution
c1 =
−1
2e
and c2 =
3e
2
So the solution to the initial value problem is
y(t) =
−1 t/2 3e −t/2
e + e
.
2e
2
We can now solve a the differential equation ay ′′ + by ′ + cy = 0, so long as the
characteristic equation has two different real roots. But this is not always the case.
We will have to deal with the possibility of having either a repeated root or complex
roots to the characteristic equation. We will deal with repeated roots now, and leave
complex roots for next time.
7
Repeated Roots: Reduction of Order
Consider a second order equation which has a characteristic equation with one repeated root. We know from theory of complex numbers that the repeated root is real,
since complex roots appear in conjugate pairs.
We want to deal with the case where we have a repeated root to the characteristic
equation, so we can simplify the characteristic equation to the form
(r − a)2 = 0
or
r 2 − 2ar + a2 = 0.
So the differential equation must be of the form
y ′′ − 2ay ′ + a2 y = 0
(6)
We know one solution is eat . We know from the principle of superposition that ceat
is a solution for any constant c, so let us try to extend this idea and find a function
u(t) such that u(t)eat is a solution to equation (6).
The only way to determine if ueat is a solution is to plug it into equation (6) and see
if it works, so we need to calculate y ′ and y ′′ in the case where y = ueat :
y ′ = (ueat )′ = uaeat + u′ eat
y ′′ = (uaeat + u′ aeat )′
= ua2 eat + u′ aeat + u′ aeat + u′′ eat
= ua2 eat + 2u′ aeat + u′′ eat
Now we plug into (6), and regroup the terms according to the order of the derivative
of u:
y ′′ − 2ay ′ + a2 y =
h
i
h
i
ua2 eat + 2u′ aeat + u′′ eat − 2a uaeat + u′eat + a2 ueat
i
h
h
= u′′ eat + u′ 2aeat − 2aeat + u a2 eat − 2a(aeat ) + a2 eat
= u′′ eat
i
To have y be a solution, we then need to have u′′ eat = 0. Since eat 6= 0, we must have
u′′ = 0, so it is easy to see that u(t) must be a linear function:
u(t) = c1 + c2 t
where c1 and c2 are any constants. Thus, our second solution is u(t)eat = c1 eat +c2 teat .
(The technique above is called reduction of order. In general, it results in a differential
equation involving u′′ and u′ which can be solved for the unknown function u′ , and
then finally for u, in cases where we know one solution to a differential equation and
hope to find a second.)
Thus we could try using eat and c1 eat + c2 teat as a fundamental set of solutions, but
it is simpler to use linear combinations of these and take
y1 (t) = eat
y2 (t) = teat
as our fundamental set of solutions.
Of course, to know that these are really a fundamental set requires checking the
Wronskian to see if these are linearly independent:
eat
teat
aeat ateat + eat
= aeat (ateat + eat ) − teat aeat = eat 6= 0.
So in the case where we have a repeated root a to the characteristic equation, a
fundamental set of solutions is given by
y1 (t) = eat
y2 (t) = teat
Example:
Find the general solution to 9y ′′ − 42y ′ + 49y = 0.
The characteristic equation is
9r 2 − 42r + 49 = (3r − 7)(3r − 7) = 0
so we have a repeated solution r = 7/3. Therefore, we have fundamental solutions
y1 (t) = e(7/3)t
and y2 (t) = te(7/3)t
and the general solution is
y(t) = c1 e(7/3)t + c2 te(7/3)t .
We can now solve constant coefficient equations if the characteristic equation has two
distinct real roots or if it has one repeated real root. We will deal with complex roots
next time.