general solution of linear differential equation

general solution of linear differential equation

The

general solution

y

of the

nonhomogeneous

linear differential equation

d

n

⁢

y

d

⁢

x

n

+

P

1

⁢

(

x

)

⁢

d

n

-

1

⁢

y

d

⁢

x

n

-

1

+

⋯

+

P

n

-

1

⁢

(

x

)

⁢

d

⁢

y

d

⁢

x

+

P

n

⁢

(

x

)

⁢

y

=

Q

⁢

(

x

)

is gotten by adding the general solution

y

¯

of the corresponding

homogeneous equation

d

n

⁢

y

¯

d

⁢

x

n

+

P

1

⁢

(

x

)

⁢

d

n

-

1

⁢

y

¯

d

⁢

x

n

-

1

+

⋯

+

P

n

-

1

⁢

(

x

)

⁢

d

⁢

y

¯

d

⁢

x

+

P

n

⁢

(

x

)

⁢

y

¯

=

0

to some particular solution of the nonhomogeneous equation.

The general solution of the homogeneous equation has the form

y

¯

=

C

1

⁢

y

1

+

C

2

⁢

y

2

+

⋯

+

C

n

⁢

y

n

(1)

where

y

1

,

y

2

,

…

,

y

n

are

linearly independent

solutions of the equation.  A particular solution of the nonhomogeneous equation can be obtained by using the method of

variation of constants

C

1

,

C

2

,

…

,

C

n

in (1).

Example 1.

Find the general solution of the nonhomogeneous linear second order

differential equation

d

2

⁢

y

d

⁢

x

2

-

4

⁢

y

=

e

x

.

(2)

The corresponding homogeneous equation

d

2

⁢

y

¯

d

⁢

x

2

=

4

⁢

y

¯

has apparently the linearly independent solutions

y

¯

=

e

±

2

⁢

x

and thus the general solution

y

¯

=

C

1

⁢

e

2

⁢

x

+

C

2

⁢

e

-

2

⁢

x

.  For finding a particular solution of (2) we variate the constants

C

1

,

C

2

, i.e. think that

C

1

=

C

1

⁢

(

x

)

,

C

2

=

C

2

⁢

(

x

)

in the sum

y

=

C

1

⁢

e

2

⁢

x

+

C

2

⁢

e

-

2

⁢

x

.

The

first derivative

y

′

=

[

C

1

′

⁢

e

2

⁢

x

+

C

2

′

⁢

e

-

2

⁢

x

]

+

[

2

⁢

C

1

⁢

e

2

⁢

x

-

2

⁢

C

2

⁢

e

-

2

⁢

x

]

of it reduces to
the latter bracket

expression

if we set the condition

C

1

′

⁢

e

2

⁢

x

+

C

2

′

⁢

e

-

2

⁢

x

=

0

.

(3)

So the second derivative is

y

′′

=

(

2

⁢

C

1

′

⁢

e

2

⁢

x

-

2

⁢

C

2

′

⁢

e

-

2

⁢

x

)

+

(

4

⁢

C

1

⁢

e

2

⁢

x

+

4

⁢

C

2

⁢

e

-

2

⁢

x

)

.

Substituting this and the expression of

y

in the differential equation (2) gives the equation

2

⁢

C

1

′

⁢

e

2

⁢

x

-

2

⁢

C

2

′

⁢

e

-

2

⁢

x

=

e

x

.

(4)

Now we have the pair of

linear equations

formed by (3) and (4) for determining the

derivatives

C

1

′

and

C

2

′

; the result of them is

C

1

′

=

e

-

x

/

4

,

C

2

′

=

-

e

3

⁢

x

/

4

.

If we then

integrate

and chose

C

1

=

-

e

-

x

/

4

,

C

2

=

-

e

3

⁢

x

/

12

,

we can form the particular solution

y

=

-

e

-

x

4

⁢

e

2

⁢

x

-

e

3

⁢

x

12

⁢

e

-

2

⁢

x

≡

-

e

x

3

.

Accordingly, the general solution of the nonhomogeneous equation (2) is

y

=

-

e

x

3

+

C

1

⁢

e

2

⁢

x

+

C

2

⁢

e

-

2

⁢

x

.

In some cases it is not necessary to use the

variation of parameters

method

above illustrated, but a particular solution may be found at simple sight, as it is the case in the following example about boundary values.

Example 2.

Find the general solution of the nonhomogeneous linear second order differential equation

y

′′

-

y

=

2

⁢

x

under the

boundary conditions

y

⁢

(

1

)

=

0

,

y

′

⁢

(

0

)

=

0

.

The function

x

↦

-

2

⁢

x

is evidently a particular solution of the differential equation. Therefore, the general solution is

y

⁢

(

x

)

=

-

2

⁢

x

+

C

1

⁢

e

x

+

C

2

⁢

e

-

x

.

Thus we have

y

′

⁢

(

x

)

=

-

2

+

C

1

⁢

e

x

-

C

2

⁢

e

-

x

. By making use of the boundary conditions, we obtain

0

=

y

⁢

(

1

)

=

-

2

+

C

1

⁢

e

+

C

2

⁢

e

-

1

,

0

=

y

′

⁢

(

0

)

=

-

2

+

C

1

-

C

2

.

Solving this

system of linear equations

and introducing

C

1

and

C

2

into the general solution, we have the result

y

⁢

(

x

)

=

-

2

⁢

x

+

2

⁢

(

e

+

1

)

e

2

+

1

⁢

e

x

-

2

⁢

e

⁢

(

e

-

1

)

e

2

+

1

⁢

e

-

x

.

To solve more advanced problems about nonhomogeneous ordinary linear differential equations of second order with boundary conditions, we may find out a particular solution by using, for instance, the

Green’s function method.

Thus consider, for instance, the self-adjoint differential equation

1

1

Minus sign, on the right-hand member of the equation, it is by convenience in the applications.

d

d

⁢

x

⁢

(

p

⁢

(

x

)

⁢

d

⁢

y

d

⁢

x

)

+

q

⁢

(

x

)

⁢

y

=

-

f

⁢

(

x

)

,

a

<

x

<

b

,

y

⁢

(

a

)

=

y

⁢

(

b

)

=

0

.

The solution of this problem, about boundary values, is known to be given by

y

⁢

(

x

)

=

∫

a

b

G

⁢

(

x

,

ξ

)

⁢

f

⁢

(

ξ

)

⁢

d

⁢

(

ξ

)

,

where the symmetric function

G

⁢

(

x

,

ξ

)

=

G

⁢

(

ξ

,

x

)

2

2

Some authors call this

symmetry

reciprocity’s law

.

is the so-called Green’s function. It satifies the following boundary problem

3

3

It is easy verify the details about such statement; it can be found in any good book on

mathematical analysis

.

{

i

)

d

d

⁢

x

(

p

(

x

)

d

⁢

G

d

⁢

x

)

+

q

(

x

)

G

=

0

,

x

≠

ξ

.

i

i

)

G

(

a

)

=

G

(

b

)

=

0

,

i

i

i

)

G

(

ξ

-

)

=

G

(

ξ

+

)

=

0

,

i

v

)

d

⁢

G

d

⁢

x

(

ξ

+

)

-

d

⁢

G

d

⁢

x

(

ξ

-

)

=

-

1

p

⁢

(

ξ

)

.

From the last two one, we realize that

G

is

continuous at

x

=

ξ

while

d

⁢

G

/

d

⁢

x

has there a

jump discontinuity

.

4

4

The solution

y

⁢

(

x

)

, which is above given, it may be physically interpreted as follows: if

y

stands for a displacement and

f

like a force per length unit, then the Green’s function

G

⁢

(

x

,

ξ

)

corresponds to a displacement at

x

due a force, of unit magnitude, concentrated at

x

=

ξ

.

Let us see an example.

Example 3.

Consider the problem

{

d

2

⁢

y

d

⁢

x

2

=

-

f

⁢

(

x

)

0

<

x

<

1

,

y

⁢

(

0

)

=

y

⁢

(

1

)

=

0

.

Here,

p

⁢

(

x

)

≡

1

,

q

⁢

(

x

)

≡

0

,

a

=

0

,

b

=

1

. So from i) and ii),

d

2

⁢

G

/

d

⁢

x

2

=

0

and therefore

G

⁢

(

x

,

ξ

)

=

{

C

1

⁢

(

ξ

)

⁢

x

x

<

ξ

,

C

2

⁢

(

ξ

)

⁢

(

1

-

x

)

x

>

ξ

.

Since

ξ

stays fixed on above Green’s conditions, constants

C

1

,

C

2

may depend on

ξ

. Further, the symmetry of

G

demands that

C

1

⁢

(

ξ

)

=

C

⁢

(

1

-

ξ

)

, and

C

2

⁢

(

ξ

)

=

C

⁢

ξ

, where

C

is a constant independent on

ξ

. Then the

continuity

condition iii) is automatically satisfied, and the jump condition iv) gives

(

-

1

)

⁢

C

⁢

ξ

-

C

⁢

(

1

-

ξ

)

=

-

1

,

whence

C

=

1

.

Therefore,

G

⁢

(

x

,

ξ

)

=

{

(

1

-

ξ

)

⁢

x

x

≤

ξ

,

(

1

-

x

)

⁢

ξ

x

≥

ξ

.

Thus, the solution is

y

⁢

(

x

)

=

∫

0

x

(

1

-

x

)

⁢

ξ

⁢

f

⁢

(

ξ

)

⁢

𝑑

ξ

+

∫

x

1

x

⁢

(

1

-

ξ

)

⁢

f

⁢

(

ξ

)

⁢

𝑑

ξ

.

If, for example,

f

⁢

(

x

)

≡

1

, then we find

y

⁢

(

x

)

=

1

2

⁢

(

1

-

x

)

⁢

x

2

+

1

2

⁢

x

⁢

(

1

-

x

)

2

=

1

2

⁢

x

⁢

(

1

-

x

)

.

In some cases related to

partial differential equations

(specially that of hyperbolic type), the method of

separation of variables

, splits in

ordinary differential equations

(possibly with

variable

coefficients

) on boundary values, and one of them usually leading to a Sturm-Liouville problem (basically an eigen-values and eigen-functions problem). The general solution of those partial differential equations generally leads to Bessel-Fourier series, but the details about that question is out of the sight of this entry.

Title

general solution of linear differential equation

Canonical

name

GeneralSolutionOfLinearDifferentialEquation

Date of creation

2013-03-22 16:32:25

Last modified on

2013-03-22 16:32:25

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Result

Classification

msc 15A06

Classification

msc 34A05

Synonym

solution of linear ordinary differential equation

Related topic

LinearProblem

Related topic

SecondOrderLinearDifferentialEquation

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