620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
620123 Applied Mathematics (Advanced)
Lectures on second order (ordinary) differential equations
S. Ole Warnaar
Department of Mathematics and Statistics
The University of Melbourne
620123 Applied
Mathematics
(Advanced)
Introduction
S. Ole Warnaar
Introduction
Reduction of
order
Definition
An equation of the form
Second order
linear ODEs
d2 y
dy = f x, y ,
2
dx
dx
Second-order
linear ODEs with
constant
coefficients
is called a second-order ODE
Examples of second order ODEs are
x2
d2 y
dy
+ xy
+ sin(xy ) = 0
2
dx
dx
and
y 00 (t) = y (t).
Unfortunatelty, second order ODEs are even harder to solve than first
order ODEs and we will only look at a limited number of (very)
special cases.
620123 Applied
Mathematics
(Advanced)
Reduction of order
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Some second order ODEs are really not much more than first order
ODEs in disguise and can be reduced to the latter by reduction of
order.
We will consider two classes of such reducible second order ODEs.
The “independent” y -variable is missing:
dy d2 y
= f x,
.
2
dx
dx
The “independent” x-variable is missing:
dy d2 y
=
f
y,
.
dx 2
dx
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
When the independent y -variable is missing we simply set
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
p(x) =
dy
.
dx
This gives the following first order ODE for p:
p 0 (x) = f x, p(x) .
Solving this for p(x) and then integrating yields y (x).
Note: Since the general solution of the p-ODE contains an arbitary
constant, the general solution of the y -ODE will contain two arbitrary
constants. This is a more general feature of second order ODEs.
620123 Applied
Mathematics
(Advanced)
Example 1
S. Ole Warnaar
Find the general solution of
Introduction
x y 00 + y 0 = 6x
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Since the independent y -variable is missing we set
p = y0
so that the p-ODE reads
xp 0 + p = 6x.
This may be rewritten as
(xp)0 = 6x
so that
xp = 3x 2 + C .
The general solution of the p-ODE is thus
p(x) = 3x +
C
.
x
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
To obtain y we need one more integration:
Z
y (x) = y 0 (x) dx
Z
= p(x) dx
Z C
dx
=
3x +
x
3
= x 2 + C log|x| + D
2
for x 6= 0.
620123 Applied
Mathematics
(Advanced)
Example 2
S. Ole Warnaar
Find the general solution of the IVP
Introduction
Reduction of
order
2 −2
e x
x
x 2 y 00 − 2y 0 =
y 0 (1) = 2y (1) = − e−2
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Since the independent y -variable is missing we again set
p = y0
so that the p-ODE reads
x 2 p 0 − 2p =
2 −2
e x.
x
This is a first order linear ODE with canonical form
2
2p
2
p 0 − 2 = 3 e− x .
x
x
As integrating factor we may take
Z dx 2
I (x) = exp −2
=
exp
.
x2
x
620123 Applied
Mathematics
(Advanced)
Multiplying the canonical form by I (x) leads to
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
2 0
p ex
=
2
x3
so that the general solution of the p-ODE is
1 2
p(x) = C − 2 e− x .
x
If we try to find y by integrating p = y 0 we run into the problem that
we cannot explicitly calculate the integral
Z
2
e− x dx.
A way around this is to note that one of the initial conditions states
that p(1) = y 0 (1) = − e−2 .
Since
p(1) = C − 1 e−2
this implies that C = 0.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Therefore
Z
y (x) = −
1 −2
e x dx
x2
1 2
= − e− x + D.
2
Using the second initial condition: y (1) = − 12 e−2 shows that D = 0
so that we obtain the final solution
1 2
y (x) = − e− x .
2
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Note: Given a second order IVP one usually first tries to find the
general solution of the ODE containing two arbitrary constants. Then
the two constants are fixed by the two initial conditions.
The previous example shows that sometimes it is better to fix one of
the constants at an earlier stage to simplify difficult calculations.
620123 Applied
Mathematics
(Advanced)
When the independent x-variable is missing we set
S. Ole Warnaar
p(y ) =
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
dy
.
dx
(∗)
Then
d2 y
d dy d
dy
=
=
p(y ) = p 0 (y )
= p 0 (y )p(y ).
2
dx
dx dx
dx
dx
Hence the substitution (∗) transforms the ODE
y 00 = f y , y 0
into the first order ODE
pp 0 = f (y , p).
(∗∗)
Note: (∗∗) is a first order ODE for the function p(y ) and makes no
reference to any x-variables.
620123 Applied
Mathematics
(Advanced)
Example 1
S. Ole Warnaar
Solve the non-linear second order ODE
Introduction
(y + 1)y 00 = (y 0 )2
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Since the independent x-variable is missing we set
p = p(y ) = y 0
(∗).
Then (as before)
y 00 = pp 0
so that we obtain the reduced ODE
(y + 1)pp 0 = p 2
There are now two cases to consider.
1
The first case is p = 0. By (∗) this implies y 0 = 0 so that
y = C.
Of course we could have observed this singular solution directly
from the original ODE.
620123 Applied
Mathematics
(Advanced)
2
The second case is
(y + 1)p 0 = p.
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
This is a separable ODE so that
Z
Z
dy
dp
=
.
p
y +1
Computing the integrals yields
log|p| = log|y + 1| + E
which after exponentiating gives
p(y ) = D(y + 1).
By (∗) this implies
y 0 = D(y + 1).
This again is a separable ODE so that
Z
Z
dy
dx
=D
y +1
Computing the integrals yields the general solution
y (x) = B eDx − 1.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
y missing
x missing
Note: The singular solution
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
y =C
corresponds to the D = 0 (and B − 1 = C ) case of the general
solution
y (x) = B eDx − 1.
620123 Applied
Mathematics
(Advanced)
Second order linear ODEs
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Definition
An equation of the form
d2 y
dy
+ P(x)
+ Q(x)y = R(x)
dx 2
dx
(∗)
is called a second-order linear ODE
When R(x) = 0 the equation (∗) is homogeneous and when
R(x) 6= 0 it is inhomogeneous.
The solution to (∗) for R(x) = 0 is also known as the
complementary function of the full equation (∗).
An arbitrary function that solves (∗) is known as a particular
solution.
620123 Applied
Mathematics
(Advanced)
Lemma 1
S. Ole Warnaar
If y1 (x) and y2 (x) are solutions of
Introduction
d2 y
dy
+ P(x)
+ Q(x)y = 0
dx 2
dx
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
then so is the linear combination
y (x) = c1 y1 (x) + c2 y2 (x)
Proof.
y 00 (x)+P(x)y 0 (x) + Q(x)y (x)
= c1 y100 (x) + c2 y200 (x) + P(x) c1 y10 (x) + c2 y20 (x)
+ Q(x) c1 y1 (x) + c2 y2 (x)
h
i
= c1 y100 (x) + P(x)y10 (x) + Q(x)y1 (x)
h
i
+ c2 y200 (x) + P(x)y20 (x) + Q(x)y2 (x)
= c1 × 0 + c2 × 0
= 0.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Note: We are (of course) only interested in linear combinations of
linearly independent solutions y1 and y2 , i.e., y1 6= Cy2 .
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Lemma 2
If y1 (x) and y2 (x) are linearly independent solutions of
dy
d2 y
+ P(x)
+ Q(x)y = 0
dx 2
dx
for some interval I ⊂ R, then the general solution on I is given by the
linear combination
y (x) = c1 y1 (x) + c2 y2 (x).
Lemma 2 tells us that the problem of finding the general solution of a
second-order linear homogeneous ODE boils down to finding two
linearly independent solutions.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
The bad news is that we have no general method for finding two
such solutions.
The good news is that once we have found one solution, we can
always find a second (linearly) independent solution.
If we are lucky (or smart . . . ) enough to guess one solution we can
thus find the general solution.
Lemma 3
Let y1 (x) be a solution to
y 00 (x) + P(x)y 0 (x) + Q(x)y (x) = 0
Then a second solution may be found by the substitution
y2 (x) = u(x) y1 (x)
The above substitution is (again) known as reduction of order for
reasons that will become clear shortly.
620123 Applied
Mathematics
(Advanced)
Proof. We know that y1 solves the ODE, i.e.,
S. Ole Warnaar
y100 + Py10 + Qy1 = 0
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
(1).
(For brevity we suppress all x-dependence.)
Defining y2 = uy1 we get
y20 = u 0 y1 + uy10
y200
00
= u y1 +
2u 0 y10
(2a)
+
uy100 .
(2b)
We want y2 to solve the ODE, i.e.,
y200 + Py20 + Qy2 = 0.
Substituting (2a) and (2b) this becomes
u 00 y1 + 2u 0 y10 + u y100 + Py10 + Qy1 + Pu 0 y1 = 0.
|
{z
}
= 0 by (1)
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
We thus find that y2 is a solution if
y0 u 00 + u 0 P + 2 1 = 0
y1
Defining u 0 = v this becomes a first order linear ODE of the form
v 0 + f (x)v = 0.
and can be solved using an integration factor.
Since u(x) 6= C it is clear that y1 and y2 are linearly independent.
y0
(Possibly u 0 is constant (namely if P + 2 1 = 0) but then
y1
u = Cx + D.)
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Example
Solve the ODE
x 2 y 00 + 2xy 0 − 6y = 0
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
We “guess” the solution y1 (x) = x 2 .
To find a second linearly independent solution we use reduction of
order and set y2 = uy1 = ux 2 .
Then y20 = u 0 x 2 + 2ux and y200 = u 00 x 2 + 4u 0 x + 2u.
Substituting this in the ODE gives
x 2 (u 00 x 2 + 4u 0 x + 2u) + 2x(u 0 x 2 + 2ux) − 6ux 2 = 0.
The terms involving u cancel (as they should) and we are left with
u 00 x 4 + 6u 0 x 3 = 0
or
u 00 +
6 0
u = 0.
x
620123 Applied
Mathematics
(Advanced)
The linear first order ODE
u 00 +
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
6 0
u =0
x
can be integrated using the integrating factor
Z dx = exp(6 log x) = x 6 .
I (x) = exp 6
x
Therefore
(u 0 x 6 )0 = 0
and
u0 =
C
.
x6
One more integration gives us
u=−
A
C
+ B = 5 + B.
5x 5
x
Recalling that y2 = ux 2 we finally obtain
y2 (x) =
A
+ Bx 2 .
x3
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Since we are only interested in that part of y2 that is linearly
independent of y1 we may take
y2 (x) =
1
.
x3
Now that we have two linearly independent solution we know from
Lemma 2 that the general solution is given by
y (x) = c1 y1 (x) + c2 y2 (x) = c1 x 2 +
c2
.
x3
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Note: In our proof of Lemma 3 we reduced the problem of finding a
second solution of
y 00 (x) + P(x)y 0 (x) + Q(x)y (x) = 0
to the problem of finding the solution to the homogeneous first order
linear ODE
y0 u 00 + u 0 P + 2 1 = 0
y1
Because we know how to solve the above even when the right-hand
side is non-zero, the following generalization of Lemma 3 is
suggested.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Lemma 4
Let y1 (x) be a solution to
y 00 (x) + P(x)y 0 (x) + Q(x)y (x) = 0
Then the substitution y (x) = u(x)y1 (x) may be used to find a
solution to the inhomogeneous linear ODE
y 00 (x) + P(x)y 0 (x) + Q(x)y (x) = R(x).
Proof. The proof follows the proof of Lemma 3 mutatis mutandis
and lead to the ODE
y0 u 00 + u 0 P + 2 1 = R(x)
y1
for the function u.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Example
Solve the ODE
x 2 y 00 + 2xy 0 − 6y = 12x
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
From our previous example we already know two solutions to the
homogeneous ODE and we can either take
1
.
x3
We will apply Lemma 4 for either choice of y1 .
y1 (x) = x 2
or y1 (x) =
y1 = x 2
In this case y (x) = u(x)x 2 .
By almost of a repeat of our earlier calculations we get
x 2 (u 00 x 2 + 4u 0 x + 2u) + 2x(u 0 x 2 + 2ux) − 6ux 2 = 12x.
Again cancelling the terms involving u we are left with
u 00 x 4 + 6u 0 x 3 = 12x
or
u 00 +
6 0
12
u = 3.
x
x
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
By the same integrating factor as before (I (x) = x 6 )
Introduction
(u 0 x 6 )0 = 12x 3
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
and
u0 =
C
3
+ 2.
6
x
x
One more integration gives us
u=
3
B
+A− .
x5
x
Recalling that y = ux 2 we finally obtain
y (x) =
A
+ Bx 2 − 3x.
x3
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
y1 = x −3
Introduction
In this case y (x) = u(x)x −3 .
Reduction of
order
Then
Second order
linear ODEs
y 0 = u 0 x −3 − 3ux −4
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
y 00 = u 00 x −3 − 6u 0 x −4 + 12ux −5 .
Substituting this in the ODE gives
x 2 (u 00 x −3 −6u 0 x −4 +12ux −5 )+2x(u 0 x −3 −3ux −4 )−6ux −3 = 12x.
The terms involving u cancel (as they should) and we are left
with
u 00 x −1 − 4u 0 x −2 = 12x.
or
u 00 − 4u 0 x −1 = 12x 2 .
620123 Applied
Mathematics
(Advanced)
An integrating factor is
S. Ole Warnaar
Z dx = x −4 .
I (x) = exp −4
x
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Hence
(u 0 x −4 )0 = 12x −2
and
u 0 = Cx 4 − 12x 3 .
One more integration yields
u = Ax 5 + B − 3x 4 .
Since y = ux −3 this finally gives
y (x) = Ax 2 +
as before.
B
− 3x
x3
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
It may be shown that
y (x) = Ax 2 +
B
− 3x
x3
corresponds to the general solution of the linear ODE
x 2 y 00 + 2xy 0 − 6y = 12x
(∗)
so that we indeed have “solved the ODE”.
Note that x 2 and x −3 correspond to two linearly independent
solutions of the homogeneous equation. According to Lemma 2
yH (x) = Ax 2 +
B
x3
is therefore the complementary function of (∗).
Similar to the terminology used for first order linear ODEs we
call yP (x) = −3x a particular solution of (∗).
We therefore have that the general solution of (∗) is of the form
y (x) = yH (x) + yP (x).
620123 Applied
Mathematics
(Advanced)
The previous example is typical and we have the following result
S. Ole Warnaar
Introduction
Reduction of
order
Theorem 1
The general solution of the second order linear ODE
d2 y
dy
+ P(x)
+ Q(x)y = R(x)
dx 2
dx
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
is given by
y (x) = yH (x) + yP (x).
In the theorem yH (x) refers to the complementary function, i.e., to
the general solution of the ODE with R(x) = 0, and yP (x) refers to
any particular solution.
From Lemma 2 we know that yH (x) is of the form
yH (x) = c1 y1 (x) + c2 y2 (x)
with y1 and y2 two linearly independent solutions to the
homogeneous equation.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
linear
independence
Reduction of
order
Second-order
linear ODEs with
constant
coefficients
Due to the lack of a general method for finding such a solution our
best hope is to
1
Guess (or ask someone really clever) one solution y1 (x) to the
homogeneous equation.
2
Apply the substitution y (x) = y1 (x)u(x) to find the general
solution of the inhomogeneous equation.
The above is exactly what we did in the previous example.
There is one special type of second order linear ODEs that we can
solve without resorting to guess-work or phoning a clever friend . . .
620123 Applied
Mathematics
(Advanced)
Second-order linear ODEs with constant coeffients
S. Ole Warnaar
Introduction
Reduction of
order
Definition
Second order
linear ODEs
A second-order linear ODE with constant coefficients is an equation
of the form
dy
d2 y
+ c y = R(x)
(∗)
a 2 +b
dx
dx
where a, b, c are constants such that a 6= 0
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Our general method of solution is as follows.
1
2
Find the complementary function yH , i.e., the solution to (∗)
with R(x) = 0.
Find a particular solution yP .
Note: To find yH there is a method that always works, so that we do
no longer need to rely on guesswork or a clever friend.
620123 Applied
Mathematics
(Advanced)
Consider the homogeneous linear ODE with constant coefficients:
S. Ole Warnaar
Introduction
a
Reduction of
order
Second order
linear ODEs
From Lemma 2 we know that the general solution is of the form
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
d2 y
dy
+b
+ c y = 0.
dx 2
dx
y (x) = c1 y1 (x) + c2 y2 (x)
(H)
with y1 and y2 two linearly independent solutions.
To find these we try
y (x) = eλx .
Then
y 0 (x) = λ eλx
and y 00 (x) = λ2 eλx .
Substituting this in (H) yields
aλ2 eλx + bλ eλx + c eλx = 0.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Since eλx 6= 0 we may divide by eλx to obtain the characteristic
equation
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
aλ2 + bλ + c = 0
(C )
There are several possible scenarios.
1
If b 2 − 4ac > 0 then (C ) has two distinct real roots λ1 and λ2 .
Hence two linearly independent solutions are
y1 (x) = eλ1 x
and y2 (x) = eλ2 x .
Therefore, by Lemma 2,
y (x) = c1 eλ1 x + c2 eλ2 x .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
2
If b 2 − 4ac = 0 then (C ) has one real root λ1 of multiplicity two.
Hence one solution of (H) is given by
Reduction of
order
y1 (x) = eλ1 x .
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
According to Lemma 3 we can find a second, linearly
independent solution by reduction of order, i.e., by setting
y2 = uy1 . Doing the algebra (homework) this leads to
y2 (x) = x eλ1 x .
Note: You can also directly verify that this indeed solves (H)
(provided that b 2 − 4ac = 0.)
By Lemma 2
y (x) = c1 eλ1 x + c2 x eλ1 x = (c1 + c2 x) eλ1 x .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
3
If b 2 − 4ac < 0 then (C ) has a pair of complex conjugate roots
λ1 = α + iβ and λ2 = α − iβ with β 6= 0.
Hence two linearly independent solutions are
Reduction of
order
y1 (x) = eαx+iβx
and y2 (x) = eαx−iβx .
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Therefore, by Lemma 2,
y (x) = c1 eαx+iβx + c2 eαx−iβx
= c1 eiβx + c2 e−iβx eαx
Homogeneous
Inhomogeneous
IVP
nth order
Since
eiθ = cos θ + i sin θ
We may also write the above solution as
y (x) = b1 cos(βx) + b2 sin(βx) eαx
(where b1 = c1 + c2 and b2 = i(c1 − c2 )).
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Example 1
Solve
y 00 + 7y 0 + 12y = 0
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
The characteristic equation is
λ2 + 7λ + 12 = 0.
This has two distinct roots:
λ1 = −3
and λ2 = −4.
Two linearly independent solutions are thus
y1 (x) = e−3x
and y2 (x) = e−4x
and the general solution is
y (x) = c1 e−3x + c2 e−4x .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Example 2
Solve
y 00 + 2y 0 + y = 0
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
The characteristic equation is
λ2 + 2λ + 1 = 0.
This has a single root of multiplicity two:
λ1 = −1.
Two linearly independent solutions are thus
y1 (x) = e−x
and y2 (x) = x e−x
and the general solution is
y (x) = (c1 + c2 x) e−x .
620123 Applied
Mathematics
(Advanced)
Example 3
S. Ole Warnaar
Solve
y 00 − 4y 0 + 5y = 0
Introduction
Reduction of
order
The characteristic equation is
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
λ2 − 4λ + 5 = 0.
Since
λ2 − 4λ + 5 = (λ − 2)2 + 1
this give the complex conjugate pair
λ1 = 2 + i
and λ2 = 2 − i.
Two linearly independent solutions are thus
y1 (x) = e2x cos x
and y2 (x) = e2x sin x
and the general solution is
y (x) = c1 cos x + c2 sin x) e2x .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Now that we know how to solve any homogeneous second order
linear ODE with constant coefficients we consider several examples of
the inhomogeneous counterpart
a
d2 y
dy
+b
+ c y = R(x).
dx 2
dx
(∗)
620123 Applied
Mathematics
(Advanced)
Example 1
S. Ole Warnaar
Solve
y 00 + y 0 − 2y = x 2 − 2x + 3
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
First we solve the homogeneous equation
y 00 + y 0 − 2y = 0
with corresponding characteristic equation
λ2 + λ − 2 = 0.
This has two distinct roots:
λ1 = −2
and λ2 = 1.
Two linearly independent solutions are thus
y1 (x) = e−2x
and y2 (x) = ex
and the complementary function is
yH (x) = c1 e−2x + c2 ex .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
According to Lemma 4 we can now find the solution to the
inhomogeneous equation by reduction of order, i.e., by taking
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
y (x) = u(x) e−2x
or y (x) = u(x) ex .
Let us take the second option. Then
y 0 = (u 0 + u) ex
y 00 = (u 00 + 2u 0 + u) ex .
Substituting this into the ODE gives
u 00 + 3u 0 = e−x (x 2 − 2x + 3).
This is a linear first order ODE for u 0 with integrating factor
I (x) = e3x .
Hence
0
u 0 e3x = (x 2 − 2x + 3) e2x .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Integrating once this becomes
Z
1
0 3x
u e = (x 2 − 2x + 3) e2x dx = (2x 2 − 6x + 9) e2x + C
4
so that
u0 =
1
(2x 2 − 6x + 9) e−x + C e−3x
4
Integrating a second time we find
Homogeneous
Inhomogeneous
IVP
nth order
u=
1
(−2x 2 + 2x − 7) e−x + c1 e−3x + c2 .
4
Since y = u ex we have thus found the general solution
1
y (x) = c1 e−2x + c2 ex + (−2x 2 + 2x − 7) = yH (x) + yP (x),
4
with
yP (x) =
a particular solution.
1
(−2x 2 + 2x − 7)
4
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
The previous method for finding the general solution is somewhat
cumbersome and a better method to find yP (x) proceeds as follows.
Take the ODE
y 00 + y 0 − 2y = x 2 − 2x + 3
(∗)
and observe that the function R(x) = x 2 − 2x + 3 is a polynomial.
Differentiating the ODE gives
y 000 + y 00 − 2y 0 = 2x − 2
(∗∗)
and differentiating again gives
y (4) + y (3) − 2y 00 = 2.
(∗ ∗ ∗)
Because the right-hand side is now a constant (in this case 2) it is
easy to find a particular solution to (∗ ∗ ∗), namely
yP00 = −1.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Substituting this in (∗∗) yields
Introduction
(0) + (−1) − 2yP0 = 2x − 2
Reduction of
order
Second order
linear ODEs
so that
1
yP0 = −x + .
2
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Substituting this in (∗) yields
1
− 2yP = x 2 − 2x + 3
(−1) + −x +
2
so that
yP =
1
(−2x 2 + 2x − 7).
4
Hence we have recovered the previous particular solution.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Note: The above method for finding a particular solution to a second
order ODE with constant coefficients always works, provided the
function R(x) is a polynomial.
Moreover, if the degree of the polynomials is n, then we need to
differentiate the ODE n times.
In Example 1, R(x) = x 2 − 2x + 3 which is a polynomials of degree
2. Hence we differentiated twice.
The following examples show that with a few extra tricks we may
also apply the previous method when
R(x) is a polynomial times an exponential,
R(x) is a polynomial times a simple trigonometric function.
620123 Applied
Mathematics
(Advanced)
Example 2
S. Ole Warnaar
Solve
y 00 − y = 3x e2x
Introduction
Reduction of
order
First we solve the homogeneous equation
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
y 00 − y = 0
with corresponding characteristic equation
λ2 − 1 = 0.
This has two distinct roots:
λ1 = −1
and λ2 = 1.
Two linearly independent solutions are thus
y1 (x) = e−x
and y2 (x) = ex
and the complementary function is
yH (x) = c1 e−x + c2 ex .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
It is homework to find the general solution by means of Lemma 4,
and below we take the alternative approach of finding a particular
solution without using the complementary function.
Since the right-hand side (i.e., the function R(x)) of
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
y 00 − y = 3x e2x
(∗)
is not a polynomial, no amount of differentiating will reduce it to a
constant.
We thus have to get rid of the exponential first.
The trick is to put
y (x) = u(x) e2x
with u(x) an unknown function.
Then
y 0 = (u 0 + 2u) e2x
y 00 = (u 00 + 4u 0 + 4u) e2x .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Substituting this in (∗) and dividing by e2x yields
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
u 00 + 4u 0 + 3u = 3x.
(∗∗)
Note that this only has a polynomial on the right!
To now find a particular solution of (∗∗) we differentiate to get
u 000 + 4u 00 + 3u 0 = 3
which has the particular solution uP0 = 1.
Substituting this in (∗) gives uP = x − 4/3.
Recalling that y = u e2x we thus find the following particular solution
to (∗):
4 2x
yP (x) = x −
e .
3
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Finally, from Theorem 1 we know that the general solution is given by
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
y (x) = yH (x) + yP (x).
Therefore
4 2x
y (x) = c1 e−x + c2 ex + x −
e .
3
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Example 3
Solve
y 00 − y = 2x cos x
The complementary function is the same as that of Example 2:
yH (x) = c1 e−x + c2 ex .
It is again homework to find the general solution by means of Lemma
4, and below we (again) take the alternative approach of finding a
particular solution without using the complementary function.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
From Example 2 we have learned how to handle a polynomial times
an exponential function.
This may also be used in the present case by going complex.
That is, we first determine a particular solution of
z 00 − z = 2x eix
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
(∗)
and then take the real part of this to get a particular solution of
y 00 − y = 2x cos x.
Homogeneous
Inhomogeneous
IVP
nth order
Using the previous trick we put
z(x) = u(x) eix
with u(x) an unknown function.
Then
z 0 = (u 0 + iu) eix
z 00 = (u 00 + 2iu 0 − u) eix .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Substituting this in (∗) and dividing by eix yields
u 00 + 2iu 0 − 2u = 2x.
(∗∗)
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Again this only has a polynomial on the right.
To now find a particular solution of (∗∗) we differentiate to get
u 000 + 2iu 00 − 2u 0 = 2
which has the particular solution uP0 = −1.
Substituting this in (∗∗) gives uP = −x − i.
Recalling that z = u eix we thus find the following particular solution
to (∗):
zP (x) = −(x + i) eix .
As a final step we need to take the real part to obtain a particular
solution yP (x).
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Now
−(x + i) eix = −(x + i)(cos x + i sin x)
Second order
linear ODEs
= (sin x − x cos x) + i(. . . ).
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Therefore
yP (x) = sin x − x cos x.
Again appealing to Theorem 1 yields the general solution
y (x) = c1 e−x + c2 ex + sin x − x cos x.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
We have now seen how to find a particular solution of
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
a
dy
d2 y
+b
+ c y = R(x)
dx 2
dx
without using the complementary function when R(x) is
a polynomial
a polynomial times an exponential function
a polynomial times a simple trigonometric function.
Homogeneous
Inhomogeneous
IVP
nth order
The above three cases are enough to also deal with linear
combination of the above three types, for example
R(x) = x 2 + 2x + 1 + e2x (x + 1)
or
R(x) = x 3 + cos(2x) + x ex .
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Lemma 5
Reduction of
order
If yP1 is a particular solution of
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
a
d2 y
dy
+b
+ c y = R1 (x)
dx 2
dx
and yP2 is a particular solution of
a
dy
d2 y
+b
+ c y = R2 (x)
2
dx
dx
then yP = yP1 + yP2 is a particular solution of
a
d2 y
dy
+b
+ c y = R1 (x) + R2 (x).
dx 2
dx
Proof. Homework
620123 Applied
Mathematics
(Advanced)
Example
S. Ole Warnaar
Find a particular solution of
y 00 − 2y 0 + y = x 2 + 1 + e−2x + cos x
Introduction
Reduction of
order
We first find a particular solution of
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
y 00 − 2y 0 + y = x 2 + 1.
(∗)
Hence we differentiate twice to get
Homogeneous
Inhomogeneous
IVP
nth order
y 000 − 2y 00 + y 0 = 2x
(∗∗)
and
y (4) − 2y 000 + y 00 = 2.
Therefore yP001 = 2, so that, by (∗∗)
−4 + yP0 1 = 2x.
This yields yP0 1 = 2x + 4, so that by (∗)
2 − 2(2x + 4) + yP1 = x 2 + 1.
This finally yields yP1 = x 2 + 4x + 7.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Next we find a particular solution of
y 00 − 2y 0 + y = e−2x .
Hence we set y (x) = u(x) e−2x to get
u 00 − 6u 0 + 9u = 1.
This obviously has the particular solution uP2 = 1/9 leading to
yP2 =
1 −2x
e .
9
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Finally we find a particular solution of
y 00 − 2y 0 + y = cos x.
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Hence we first consider the complex ODE
z 00 − 2z 0 + z = eix .
If we set z(x) = u(x) eix we get
u 00 + 2u 0 (i − 1) − 2iu = 1.
This obviously has the particular solutotion uP3 = i/2 leading to
zP3 =
i ix
e .
2
Taking the real part yields
1
yP3 = − sin x.
2
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
According to (an iterated version of) Lemma 5 a particular solution
of the ODE is thus
yP (x) = yP1 (x) + yP2 (x) + yP3 (x)
1
1
= x 2 + 4x + 7 + e−2x − sin x.
9
2
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Now that we know how to solve second order linear ODEs with
constant coefficients we can of course also deal with IVPs of this
type.
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Example
Solve the IVP
y 00 + y = cos x
y (0) = 2
y 0 (0) = −3
First we solve the homogeneous equation
y 00 + y = 0
with corresponding characteristic equation
λ2 + 1 = 0.
This has the pair of complex conjugate roots
λ1 = −i
and λ2 = i.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Two linearly independent solutions are thus
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
y1 (x) = e−ix
and y2 (x) = e−ix
(∗)
y1 (x) = cos x
and y2 (x) = sin x.
(∗∗)
but also
You are of course free to choose either (∗) or (∗∗).
Taking the second pair, the complementary function is
yH (x) = c1 cos x + c2 sin x.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
For the sake of variation let us use Lemma 4 to find the general
solution.
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
It is homework to find the particular solution without using the
complementary function. For this problem this is actually the quicker
method.
We set y (x) = u(x) sin x.
(It is homework to try y (x) = u(x) cos x.)
Then
y 0 = u 0 sin x + u cos x
y 00 = u 00 sin x + 2u 0 cos x − u sin x.
Substituting this in the ODE gives the first order linear ODE (in u 0 )
u 00 sin x + 2u 0 cos x = cos x.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
The canonical form is
Introduction
u 00 + 2u 0 cotan x = cotan x
Reduction of
order
Second order
linear ODEs
leading to the integrating factor
Second-order
linear ODEs with
constant
coefficients
Z
I (x) = exp 2 cotan x dx
= exp 2 log|sin x|
Homogeneous
Inhomogeneous
IVP
nth order
= sin2 x.
Therefore
(u 0 sin2 x)0 = cos x sin x
and
0
2
u sin x =
Z
sin x cos x dx =
1 2
sin x − c1 .
2
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Dividing by sin2 x this yields
Introduction
u0 =
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
1
c1
− 2 .
2 sin x
Hence
1
x + c1 cotan x + c2 .
2
Since y = u sin x we finally get
u=
y (x) =
1
x sin x + c1 cos x + c2 sin x.
2
To fix c1 and c2 we use the initial conditions to find c1 = 2 and
c2 = −3 so that
y (x) =
1
x sin x + 2 cos x − 3 sin x.
2
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Our final example shows that nearly everything we have done so far
extends to nth order linear ODEs with constant coefficients.
Example
Solve the fourth order ODE
y (4) + y (3) − y 00 + y 0 − 2y = x
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
First we solve the homogeneous equation
y (4) + y (3) − y 00 + y 0 − 2y = 0
with corresponding characteristic equation
λ4 + λ3 − λ2 + λ − 2 = 0.
Since
λ4 + λ3 − λ2 + λ − 2 = (λ − 1)(λ + 2)(λ2 + 1)
this has four distinct roots
λ1 = 1,
λ2 = −2,
λ3 = −i,
λ4 = i.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
Reduction of
order
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
Four linearly independent solutions are thus
y1 (x) = ex ,
y2 (x) = e−2x ,
y3 (x) = e−ix ,
y4 (x) = eix
but also
y1 (x) = ex ,
y2 (x) = e−2x ,
y3 (x) = cos x,
y4 (x) = sin x.
Choosing the latter, the complementary function may be written as
yH (x) = c1 ex + c2 e−2x + c3 cos x + c4 sin x.
620123 Applied
Mathematics
(Advanced)
S. Ole Warnaar
Introduction
To find a particular solution we differentiate the ODE:
Reduction of
order
y (5) + y (4) − y 000 + y 00 − 2y 0 = 1
Second order
linear ODEs
Second-order
linear ODEs with
constant
coefficients
Homogeneous
Inhomogeneous
IVP
nth order
so that yP0 = − 12 .
Substituting this in the original ODE yields
1
yP (x) = − (2x + 1)
4
so that
1
y (x) = c1 ex + c2 e−2x + c3 cos x + c4 sin x − (2x + 1).
4