on inhomogeneous second-order linear ODE with constant coefficients

on inhomogeneous second-order linear ODE with constant coefficients

Let’s consider solving the ordinary second-order

linear
differential equation

d

2

⁢

y

d

⁢

x

2

+

a

⁢

d

⁢

y

d

⁢

x

+

b

⁢

y

=

R

⁢

(

x

)

(1)

which is
inhomogeneous (

http://planetmath.org/HomogeneousLinearDifferentialEquation

), i.e.

R

⁢

(

x

)

≢

0

.

For obtaining the

general solution

of (1) we have to add to the
general solution of the
corresponding

homogeneous equation

(

http://planetmath.org/SecondOrderLinearODEWithConstantCoefficients

)

d

2

⁢

y

d

⁢

x

2

+

a

⁢

d

⁢

y

d

⁢

x

+

b

⁢

y

=

0

(2)

some
particular solution (

http://planetmath.org/SolutionsOfOrdinaryDifferentialEquation

)
of the inhomogeneous equation (1).  A latter one can
always be gotten by means of the

variation of parameters

, but
in many cases there exist simpler ways to find a particular
solution of (1).

1

∘

:

R

⁢

(

x

)

is a nonzero constant function

x

↦

c

.  In this case, apparently

y

=

c

b

is
a solution of (1), supposing that

b

≠

0

.  If

b

=

0

but

a

≠

0

,  a particular solution is

y

=

c

a

⁢

x

.
If

a

=

b

=

0

,  a solution is gotten via two consecutive
integrations.

2

∘

:

R

⁢

(

x

)

is a polynomial function of degree

n

≥

1

.  Now (1) has as solution a

polynomial

which can be
found by using indetermined coefficients.  If

b

≠

0

,
the polynomial is of degree

n

and is uniquely determined.
If

b

=

0

and

a

≠

0

,  the degree of the polynomial
is

n

+

1

and its

constant term

is arbitrary. If

a

=

b

=

0

the polynomial is of degree

n

+

2

and is
gotten via two integrations.

3

∘

:  Let

R

⁢

(

x

)

in (1) be of the form

α

⁢

sin

⁡

n

⁢

x

+

β

⁢

cos

⁡

n

⁢

x

with

α

,

β

,

n

constants.  We try to find a solution of the same form and put
into (1) the expression

y

:=

A

⁢

sin

⁡

n

⁢

x

+

B

⁢

cos

⁡

n

⁢

x

.

(3)

Then the left hand side of (1) attains the form

[

(

b

-

n

2

)

⁢

A

-

a

⁢

n

⁢

B

]

⁢

sin

⁡

n

⁢

x

+

[

a

⁢

n

⁢

A

+

(

b

-

n

2

)

⁢

B

]

⁢

cos

⁡

n

⁢

x

.

This must equal

R

⁢

(

x

)

, i.e. we have the conditions

(

b

-

n

2

)

⁢

A

-

a

⁢

n

⁢

B

=

α

and

a

⁢

n

⁢

A

+

(

b

-

n

2

)

⁢

B

=

β

.

These determine uniquely the values of

A

and

B

provided that
the

determinant

|

b

-

n

2

-

a

⁢

n

a

⁢

n

b

-

n

2

|

=

a

2

⁢

n

2

+

(

b

-

n

2

)

2

does not vanish.  Then we obtain the particular solution (3).
The determinant

vanishes

only if

a

=

0

and

b

=

n

2

, in
which case the

differential equation

(1) reads

d

2

⁢

y

d

⁢

x

2

+

n

2

⁢

y

=

α

⁢

sin

⁡

n

⁢

x

+

β

⁢

cos

⁡

n

⁢

x

.

(4)

Unless we have

α

=

β

=

0

, the equation (4) has no
solution of the form (3), since

d

2

d

⁢

x

2

⁢

(

A

⁢

sin

⁡

n

⁢

x

+

B

⁢

cos

⁡

n

⁢

x

)

+

n

2

⁢

(

A

⁢

sin

⁡

n

⁢

x

+

B

⁢

cos

⁡

n

⁢

x

)

=

0

(5)

identically.  But we find easily a solution of (4) when we

differentiate

the identity (5) with respect to

n

.  Changing
the order of

differentiations

we get

d

2

d

⁢

x

2

⁢

(

A

⁢

x

⁢

cos

⁡

n

⁢

x

-

B

⁢

x

⁢

sin

⁡

n

⁢

x

)

+

n

2

⁢

(

A

⁢

x

⁢

cos

⁡

n

⁢

x

-

B

⁢

x

⁢

sin

⁡

n

⁢

x

)

=

-

2

⁢

n

⁢

A

⁢

sin

⁡

n

⁢

x

-

2

⁢

n

⁢

B

⁢

cos

⁡

n

⁢

x

.

The right hand side coincides with the right hand side of
(4) iff

-

2

⁢

n

⁢

A

=

α

and

-

2

⁢

n

⁢

B

=

β

, and thus (4)
has the solution

y

:=

-

α

2

⁢

n

⁢

x

⁢

cos

⁡

n

⁢

x

+

β

2

⁢

n

⁢

x

⁢

sin

⁡

n

⁢

x

.

4

∘

:  Let

R

⁢

(

x

)

in (1) now be

α

⁢

e

k

⁢

x

where

α

and

k

are constants.  Denote
the left hand side of (1) briefly

d

2

⁢

y

d

⁢

x

2

+

a

d

⁢

y

d

⁢

x

+

b

y

=

:

F

(

y

)

.  We seek
again a solution of the same form

A

⁢

e

k

⁢

x

as

R

⁢

(

x

)

.

First we have

F

⁢

(

A

⁢

e

k

⁢

x

)

=

A

⁢

(

k

2

+

a

⁢

k

+

b

)

⏟

f

⁢

(

k

)

⁢

e

k

⁢

x

=

A

⁢

f

⁢

(

k

)

⁢

e

k

⁢

x

.

Thus

A

can be determined from the condition

A

⁢

f

⁢

(

k

)

=

α

.  If

f

⁢

(

k

)

≠

0

,  i.e.

k

is not a root of the

characteristic equation

f

⁢

(

r

)

=

0

corresponding the
homogeneous equation (2), then we obtain the
particular solution

y

:=

α

f

⁢

(

k

)

⁢

e

k

⁢

x

=

α

k

2

+

a

⁢

k

+

b

⁢

e

k

⁢

x

of the inhomogeneous equation (1).

If

f

⁢

(

k

)

=

0

, then

e

k

⁢

x

and

A

⁢

e

k

⁢

x

satisfy the
homogeneous equation

F

⁢

(

y

)

=

0

.  Now we may start from the
identity

F

⁢

(

A

⁢

e

r

⁢

x

)

=

A

⁢

f

⁢

(

r

)

⁢

e

r

⁢

x

and differentiate it with respect to

r

.  Changing again the
order of differentiations we can write first

F

⁢

(

A

⁢

x

⁢

e

r

⁢

x

)

=

A

⁢

e

r

⁢

x

⁢

[

f

′

⁢

(

r

)

+

x

⁢

f

⁢

(

r

)

]

,

(6)

and differentiating anew,

F

⁢

(

A

⁢

x

2

⁢

e

r

⁢

x

)

=

A

⁢

e

r

⁢

x

⁢

[

f

′′

⁢

(

r

)

+

2

⁢

x

⁢

f

′

⁢

(

r

)

+

x

2

⁢

f

⁢

(

r

)

]

.

(7)

If

k

is a

simple root

of the equation

f

⁢

(

r

)

=

0

,  i.e.
if

f

⁢

(

k

)

=

0

but

f

′

⁢

(

k

)

≠

0

,  then

r

:=

k

makes the right hand side of (6) to

A

⁢

f

′

⁢

(

k

)

⁢

e

k

⁢

x

, which
equals to

R

⁢

(

x

)

=

α

⁢

e

k

⁢

x

by choosing

A

:=

α

f

′

⁢

(

k

)

.  Then we have found the
particular solution

y

:=

α

f

′

⁢

(

k

)

⁢

x

⁢

e

k

⁢

x

=

α

2

⁢

k

+

a

⁢

x

⁢

e

k

⁢

x

.

We have still to handle the case when

k

is the double root of
the equation

f

⁢

(

k

)

=

0

and thus

f

′

⁢

(

k

)

=

0

.  Putting

r

:=

k

into (7), the right hand side reduces to

A

⁢

f

′′

⁢

(

k

)

⁢

e

k

⁢

x

=

2

⁢

A

⁢

e

k

⁢

x

; this equals to

R

⁢

(

x

)

=

α

⁢

e

k

⁢

x

when choosing

A

:=

α

2

.  So we have the
particular solution

y

:=

α

2

⁢

x

2

⁢

e

k

⁢

x

of the given inhomogeneous equation.

5

∘

:  Suppose that in (1) the right hand side

R

⁢

(

x

)

is a sum of several

functions

,

d

2

⁢

y

d

⁢

x

2

+

a

⁢

d

⁢

y

d

⁢

x

+

b

⁢

y

=

R

1

⁢

(

x

)

+

R

2

⁢

(

x

)

+

…

+

R

n

⁢

(

x

)

,

(8)

and one can find a particular solution

y

i

⁢

(

x

)

for each of
the equations

d

2

⁢

y

d

⁢

x

2

+

a

⁢

d

⁢

y

d

⁢

x

+

b

⁢

y

=

R

i

⁢

(

x

)

.

Then evidently the sum

y

1

⁢

(

x

)

+

y

2

⁢

(

x

)

+

…

+

y

n

⁢

(

x

)

is a
particular solution of the equation (8).

References

1

Ernst Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset III.1

.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

Title

on inhomogeneous second-order linear ODE with constant coefficients

Canonical name

OnInhomogeneousSecondorderLinearODEWithConstantCoefficients

Date of creation

2014-03-05 16:25:57

Last modified on

2014-03-05 16:25:57

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Derivation

Classification

msc 34A05

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