The Chemistry Maths Book, Second Edition

346 Chapter 12Second-order differential equations. Constant coefficients

Application of the boundary conditions gives

y(0) 1 = 111 = 1 e

0

(d

1

1 cos 101 + 1 d

2

1 sin 1 0) 1 = 1 d

1

y(π 2 2) 1 = 121 = 1 e

π

22

(d

1

1 cos 1 π 221 + 1 d

2

1 sin 1 π 2 2) 1 = 1 d

2

e

π

22

Therefore,d

1

1 = 11 andd

2

1 = 12 e

−π 22

, and the solution of the boundary value problem is

y(x) 1 = 1 e

x

(cos 1 x 1 + 12 e

−

π

22

1 sin 1 x)

0 Exercises 17–19

EXAMPLE 12.8Solve the boundary value problem

y′′ 1 + 1 y′ 1 − 16 y 1 = 1 0, y(0) 1 = 1 1, y(x) 1 → 1 0 as x 1 → 1 ∞

The equation is that discussed in Example 12.6, with general solution

y(x) 1 = 1 c

1

e

2 x

1 + 1 c

2

e

− 3 x

The first boundary condition gives

y(0) 1 = 111 = 1 c

1

1 + 1 c

2

The second condition requires that the solution go to zero as xgoes to infinity.

The function e

− 3 x

has this property but the function e

2 x

must be excluded. The

condition therefore requires that we setc

1

1 = 10. Thenc

2

1 = 11 and the solution of the

boundary value problem is

y(x) 1 = 1 e

− 3 x

0 Exercise 20

The equationy′′ 1 + 1 ω

2

y 1 = 10

In Examples 12.6 to 12.8, the two conditions on the solution of the differential

equation are sufficient to specify a particular solution. In some cases however,

when the differential equation itself contains parameters to be determined, one or

more additional conditions are required to completely specify the solution. Several

examples of such equations are discussed in Chapter 13. One of the most important

simple examples in the physical sciences is the equation

(12.20)

whereωis a real number. The general solution is given by either (12.16) witha 1 = 10 ,

y(x) 1 = 1 c

1

e

iωx

1 + 1 c

2

e

−iωx

(12.21)

dy

dx

y

2

2

2

+=ω 0