The Chemistry Maths Book, Second Edition

12.3 The general solution 343

(iii) Complex roots

When the discriminant a

2

1 − 14 bof the characteristic quadratic equation (12.7) is

negative, the roots are a pair of complex conjugate numbers.

EXAMPLE 12.5Solve the differential equationy′′ 1 − 12 y′ 1 + 12 y 1 = 10.

The characteristic equation is

λ

2

1 − 12 λ 1 + 121 = 10

and the roots are or

λ

1

1 = 111 + 1 i, λ

2

1 = 1 λ*

1

1 = 111 − 1 i

Two particular solutions are therefore

These are linearly independent, and the general solution is

y(x) 1 = 1 c

1

e

(1+i)x

1 + 1 c

2

e

(1−i)x

= 1 e

x

(c

1

e

ix

1 + 1 c

2

e

−ix

)

This is the exponential form of the solution. A trigonometric form is obtained by

expressinge

±ix

in terms ofsin 1 xandcos 1 xby means of Euler’s formula, equations

(8.33) and (8.34),

e

±ix

1 = 1 cos 1 x 1 ± 1 i 1 sin 1 x

Then

= 1 e

x

(d

1

1 cos 1 x 1 + 1 d

2

1 sin 1 x)

whered

1

1 = 1 c

1

1 + 1 c

2

andd

2

1 = 1 i(c

1

1 − 1 c

2

)are (new) arbitrary constants.

0 Exercises 11, 12

In the general complex case, the roots of the characteristic equation (12.7) are,

from (12.8),

λ=− ±













−

aa

b

22

2

=+ +−













ecc xicc x

x

()cos()sin

12 12

yx ec xi x c xi x

x

( )=++−(cos sin ) (cos sin )













12

yx e e yx e e

x

ix

x

ix

1

1

2

1

12

() ()

() ()

== , = =

+−

λλ

λ=±−

()

1

2

248,