The Chemistry Maths Book, Second Edition

11.3 Separable equations 319

and, by the method of substitution for the left side (Chapter 6), this is identical to

(11.11). The solution of the differential equation is completed by the evaluation of

the integrals in (11.11) by the methods discussed in Chapters 5 and 6 (if possible;

otherwise numerical methods must be used) and by the application of the appropriate

initial condition. The following examples demonstrate the method. Applications of

separable differential equations are discussed in Section 11.4.

EXAMPLE 11.4Separable equations

Each of the following equations has the form (11.9) with solution (11.11). An initial

condition is given in each case.

(i)

By equation (11.10),dy 1 = 1 ax

n

1 dx, and integration gives (whenn 1 ≠ 1 − 1 )

This is the general solution. The required particular solution is obtained by

applying the initial condition, thaty 1 = 10 whenx 1 = 10. Thereforec 1 = 10 and

(ii)

Separation of the variables and integration gives

An alternative form of the general solution is obtained by taking the exponential

of each side of the equation. On the left side, e

lny

1 = 1 y; on the right side,

e

kx+c

1 = 1 e

c

1 e

kx

1 = 1 ae

kx

. Then

y 1 = 1 ae

kx

By the initial condition,y 1 = 1 y

0

whenx 1 = 1 x

0

. Therefore,y(x

0

) 1 = 1 y

01

= 1 ae

kx

0

so that

a 1 = 1 y

0

e

−kx

0

, and the particular solution is

y(x) 1 = 1 y

0

e

k(x−x

0

)

dy

y

kdx

dy

y

=,ZZ=kdx ykxc,ln=+

dy

dx

=,ky y x y() .=

00

yx

ax

n

n

()=

• 
  

+ 1

1

ZZdy a x dx y x

ax

n

c

n

n

=, =

• 
  

• 
  

+

()

1

1

dy

dx

ax y

n

=,() .00=