The Chemistry Maths Book, Second Edition

12.3 The general solution 341

and the possible solutions of type (12.6) are

(12.9)

EXAMPLE 12.3The characteristic equation of the differential equation

is

λ

2

1 − 15 λ 1 + 161 = 10

The quadratic can be factorized,

λ

2

1 − 15 λ 1 + 161 = 1 (λ 1 − 1 2)(λ 1 − 1 3)

and its roots areλ

1

1 = 12 andλ

2

1 = 13. Two particular solutions of the differential

equation are therefore

and, because these functions are linearly independent, the general solution is

y 1 = 1 c

1

y

1

1 + 1 c

2

y

2

1 = 1 c

1

e

2 x

1 + 1 c

2

e

3 x

0 Exercises 7, 8

In Example 12.3 the characteristic equation has two distinct real roots so thaty

1

and

y

2

are linearly independent and the general solution is a linear combination of them.

The nature of the roots (12.8) depends on the discriminanta

2

1 − 14 b. If aand bare real

numbers, the three possible types are

(i)a

2

1 − 14 b 1 > 10 two distinct real roots

(ii)a

2

1 − 14 b 1 = 10 one real double root

(iii)a

2

1 − 14 b 1 < 10 a pair of complex conjugate roots

(i) Two distinct real roots

The roots are given by (12.8), the particular solutions by (12.9) and, as in Example

12.3, the general solution is

yx cy x cy x ce ce (12.10)

xx

() ()=+ =+()

11 2 2 1 2

12

λλ

ye e ye e

x

x

x

x

1

2

2

3

12

==, = =

λλ

dy

dx

dy

dx

y

2

2

−+= 560

ye ye

xx

12

12

=, =

λλ