The Chemistry Maths Book, Second Edition

360 Chapter 12Second-order differential equations. Constant coefficients

This is equal to 2x

2

if

(2a

2

1 + 13 a

1

1 + 12 a

0

) 1 = 1 0, (6a

2

1 + 12 a

1

) 1 = 1 0, a

2

1 = 11

so thata

2

1 = 11 ,a

1

1 = 1 − 3 anda

0

1 = 1722. A particular solution is therefore

0 Exercise 30

The method used in this example is the method of undetermined coefficients; a

0

, a

1

and a

2

being the coefficients to be determined in this case. We consider first how,

given a particular solution, a general solution may be obtained.

Lety

p

be a particular solution of the inhomogeneous equation (12.75), so that

(12.76)

Lety

h

be the general solution (12.5) of the corresponding homogeneousequation

(12.12), called the reduced equationin this context:

y

h

(x) 1 = 1 c

1

y

1

(x) 1 + 1 c

2

y

2

(x) (12.77)

(12.78)

It follows that the sum of the functionsy

h

(x)andy

p

(x),

y(x) 1 = 1 y

h

(x) 1 + 1 y

p

(x) (12.79)

is also a solution of the inhomogeneous equation:

using equations (12.76) and (12.78). The function (12.79) contains two arbitrary

constants and is the general solution of the inhomogeneous equation. The function

y

h

is often called the complementary functionandy

p

the particular integral:

general solution 1 = 1 complementary function 1 + 1 particular integral

EXAMPLE 12.13Find the general solution of the equationy′′ 1 + 13 y′ 1 + 12 y 1 = 12 x

2

.

By Example 12.12, the particular integral is

yx xx

p

()=− +

7

2

3

2

dy

dx

a

dy

dx

by

dy

dx

a

dy

dx

by

hh

h

2

2

2

2

++= + +





















































+++=

dy

dx

a

dy

dx

by r x

pp

p

2

2

()

dy

dx

a

dy

dx

by

hh

h

2

2

++= 0

dy

dx

a

dy

dx

by r x

pp

p

2

2

++=()

yxx=− +

7

2

3

2