second-order linear ODE with constant coefficients

second-order linear ODE with constant coefficients

Let’s consider the ordinary

second-order

linear differential equation

d

2

⁢

y

d

⁢

x

2

+

a

⁢

d

⁢

y

d

⁢

x

+

b

⁢

y

=

0

(1)

which is

homogeneous

(

http://planetmath.org/HomogeneousLinearDifferentialEquation

)
and the

coefficients

a

,

b

of which are

constants

.  As
mentionned in the entry
“finding another particular solution of linear ODE”, a simple substitution
makes possible to eliminate from it the addend containing

first
derivative

of the unknown

function

.  Therefore we
concentrate upon the case

a

=

0

.  We have two cases
depending on the sign of

b

=

±

k

2

.

1

∘

.

b

>

0

.  We will solve the equation

d

2

⁢

y

d

⁢

x

2

+

k

2

⁢

y

=

0

.

(2)

Multiplicating both addends by the

expression

2

⁢

d

⁢

y

d

⁢

x

it becomes

2

⁢

d

⁢

y

d

⁢

x

⁢

d

2

⁢

y

d

⁢

x

2

+

2

⁢

k

2

⁢

y

⁢

d

⁢

y

d

⁢

x

=

0

,

where the left hand side is the

derivative

of

(

d

⁢

y

d

⁢

x

)

2

+

k

2

⁢

y

2

.  The latter one thus has a constant value
which must be nonnegative; denote it by

k

2

⁢

C

2

.  We then have the equation

(

d

⁢

y

d

⁢

x

)

2

=

k

2

⁢

(

C

2

-

y

2

)

.

(3)

After taking the square root and separating the

variables

it reads

d

⁢

y

±

C

2

-

y

2

=

k

⁢

d

⁢

x

.

Integrating (see the

table of integrals

) this yields

arcsin

⁡

y

C

=

k

⁢

(

x

-

x

0

)

where

x

0

is another constant.  Consequently, the

general
solution

of the

differential equation

(2) may be written

y

=

C

⁢

sin

⁡

k

⁢

(

x

-

x

0

)

(4)

in which

C

and

x

0

are arbitrary real constants.

If one denotes

C

⁢

cos

⁡

k

⁢

x

0

=

C

1

and

-

C

⁢

sin

⁡

k

⁢

x

0

=

C

2

,
then (4) reads

y

=

C

1

⁢

sin

⁡

k

⁢

x

+

C

2

⁢

cos

⁡

k

⁢

x

.

(5)

Here,

C

1

and

C

2

are arbitrary constants.  Because both

sin

⁡

k

⁢

x

and

cos

⁡

k

⁢

x

satisfy

the given equation (2) and are

linearly independent

, its general solution can be written as (5).

2

∘

.

b

<

0

.  An analogical treatment of
the equation

d

2

⁢

y

d

⁢

x

2

-

k

2

⁢

y

=

0

.

(6)

yields for it the general solution

y

=

C

1

⁢

e

k

⁢

x

+

C

2

⁢

e

-

k

⁢

x

(7)

(note that one can eliminate the square root from the equation

y

±

y

2

+

C

=

C

′

⁢

e

k

⁢

x

and its “inverted equation”

y

∓

y

2

+

C

=

-

C

C

′

⁢

e

-

k

⁢

x

).  The linear independence
of the

obvious

solutions

e

±

k

⁢

x

implies also the linear
independence of

cosh

⁡

k

⁢

x

and

sinh

⁡

k

⁢

x

and thus allows us to
give the general solution also in the alternative form

y

=

C

1

⁢

sinh

⁡

k

⁢

x

+

C

2

⁢

cosh

⁡

k

⁢

x

.

(8)

Remark.

The standard method for solving a
homogeneous (

http://planetmath.org/HomogeneousLinearDifferentialEquation

)
ordinary second-order linear differential equation (1) with
constant coefficients is to use in it the
substitution

y

=

e

r

⁢

x

(9)

where

r

is a constant; see the entry “

second order linear
differential equation with constant coefficients

”.  This method
is possible to use also for such equations of higher order.

References

1

Ernst Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset III.1

.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

Title

second-order linear ODE with constant coefficients

Canonical

name

SecondorderLinearODEWithConstantCoefficients

Date of creation

2014-03-01 17:02:54

Last modified on

2014-03-01 17:02:54

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Derivation

Classification

msc 34A05

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