The Chemistry Maths Book, Second Edition

342 Chapter 12Second-order differential equations. Constant coefficients

(ii) A real double root

When a

2

1 − 14 b 1 = 10 the two roots (12.8) are equal, with λ

1

1 = 1 λ

2

1 = 1 −a 22. Only one

particular solution is therefore obtained from the characteristic equation:

(12.11)

Given one particular solution of a homogeneous equation it is always possible to find

a second. In the present case, a second solution, linearly independent ofy

1

, is

y

2

(x) 1 = 1 xy

1

(x) 1 = 1 xe

−ax 22

(12.12)

This is shown to be a solution by substitution into the differential equation (12.3).

Thus

so that

= 10

sincea

2

1 − 14 b 1 = 10. The general solution in this case is therefore

y(x) 1 = 1 c

1

e

−ax 22

1 + 1 c

2

xe

−ax 22

1 = 1 (c

1

1 + 1 c

2

x)e

−ax 22

(12.13)

EXAMPLE 12.4Solve the differential equationy′′ 1 − 14 y′ 1 + 14 y 1 = 10.

The characteristic equation

λ

2

1 − 14 λ 1 + 141 = 1 (λ 1 − 1 2)

2

1 = 10

has the double rootλ 1 = 12. Two particular solutions are thereforee

2 x

andxe

2 x

, and

the general solution is

y(x) 1 = 1 (c

1

1 + 1 c

2

x)e

2 x

0 Exercises 9, 10

=− +

















=− −

( )

−−

ax

bx e

x

abe

ax ax

2

22 2

44

4

=−++−+

















−

a

ax

a

ax

bx e

ax

22

2

42

dy

dx

a

dy

dx

by a

ax

ea

ax

2

2

2

2

2

2

2

4

++=−+ 1

















+−

−

aax

ebxe

ax ax

2

22













• 
  

−−

dy

dx

ax e

dy

dx

aax e

ax ax

2

2

2

2

2

22

=− 12 4

()

,=−+

( )

−−

yx e e

x

ax

1

2

1

()==

−

λ