The Chemistry Maths Book, Second Edition

362 Chapter 12Second-order differential equations. Constant coefficients

By Table 12.1, case 1 , the choice of particular integral should bey

p

1 = 1 ke

− 2 x

, but this

is already a solution of the homogeneous equation. By prescription (a)therefore,

we use

y

p

1 = 1 kxe

− 2 x

Then

y′

p

1 = 1 ke

− 2 x

1 − 12 kxe

− 2 x

, y

p

′′ 1 = 1 − 4 ke

− 2 x

1 + 14 kxe

− 2 x

so that

y

p

′′ 1 + 13 y′

p

111 + 12 y

p

1 = 1 −ke

− 2 x

and this is equal toy

p

1 = 13 e

− 2 x

ifk 1 = 1 − 3. Theny

p

1 = 13 xe

− 2 x

and the general solution is

y(x) 1 = 1 y

h

(x) 1 + 1 y

p

(x) 1 = 1 c

1

e

−x

1 + 1 c

2

e

− 2 x

1 − 13 xe

− 2 x

0 Exercises 33, 34

EXAMPLE 12.15Find the general solution of the equationy′′ 1 − 14 y′ 1 + 14 y 1 = 1 e

2 x

.

By Example 12.4, the characteristic equation of the homogeneous equation has the

double rootλ 1 = 12 , and the complementary function is

y

h

(x) 1 = 1 (c

1

1 + 1 c

2

x)e

2 x

In this case, by prescription (b), the functiony

p

1 = 1 ke

2 x

is multiplied byx

2

. Then

y

p

1 = 1 kx

2

e

2 x

, y′

p

1 = 12 kxe

2 x

1 + 12 kx

2

e

2 x

, y

p

′′ 1 = 12 ke

2 x

1 + 18 kxe

2 x

1 + 14 kx

2

e

2 x

so that

y

p

′′ 1 − 14 y′

p

111 + 14 y

p

1 = 12 ke

2 x

and this is equal toe

2 x

whenk 1 = 1122. The general solution is therefore

0 Exercise 35

0 Exercises 36–38

yx y x y x c cx x e

hp

x

() () ()=+=++













12

22

1

2