The Chemistry Maths Book, Second Edition

12.4 Particular solutions 345

EXAMPLE 12.6Solve the initial value problem

y′′ 1 + 1 y′ 1 − 16 y 1 = 1 0, y(0) 1 = 1 0, y′(0) 1 = 15

The characteristic equation is λ

2

1 + 1 λ 1 − 161 = 1 (λ 1 − 1 2)(λ 1 + 1 3) 1 = 10 , with roots λ

1

1 = 12 and

λ

2

1 = 1 − 3. The general solution is

y(x) 1 = 1 c

1

e

2 x

1 + 1 c

2

e

− 3 x

with first derivative

y′(x) 1 = 12 c

1

e

2 x

1 − 13 c

2

e

− 3 x

Then

y(0) 1 = 101 = 1 c

1

1 + 1 c

2

, y′(0) 1 = 151 = 12 c

1

1 − 13 c

2

with solutionc

1

1 = 11 , c

2

1 = 1 − 1. The solution of the initial value problem is therefore

y(x) 1 = 1 e

2 x

1 − 1 e

− 3 x

0 Exercises 13–16

(b) Boundary conditions

In many applications in the physical sciences the variable xis a coordinate and the

physical situation is determined by conditions on the value ofy(x)at the boundary of

the system. The conditions

y(x

1

) 1 = 1 y

1

, y(x

2

) 1 = 1 y

2

(12.19)

are called boundary conditions, and x

1

and x

2

are the end-points of the interval

x

1

1 ≤ 1 x 1 ≤ 1 x

2

within which the differential equation is defined. A differential equation

with boundary conditions is called a boundary value problem.

EXAMPLE 12.7Solve the boundary value problem

y′′ 1 − 12 y′ 1 + 12 y 1 = 1 0, y(0) 1 = 1 1, y(π 2 2) 1 = 12

The differential equation is that solved in Example 12.5, with general solution (in the

trigonometric form)

y(x) 1 = 1 e

x

(d

1

1 cos 1 x 1 + 1 d

2

1 sin 1 x)