Mathematical Methods for Physics and Engineering : A Comprehensive Guide

15.2 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS

where

g(x)=a 0 (x)−^14 [a 1 (x)]^2 −^12 a′ 1 (x)

h(x)=f(x)exp

{

1

2

∫

a 1 (z)dz

}

.

Since (15.73) is of a simpler form than the original equation, (15.70), it may

prove easier to solve.

Solve

4 x^2

d^2 y

dx^2

+4x

dy

dx

+(x^2 −1)y=0. (15.74)

Dividing (15.74) through by 4x^2 , we see that it is of the form (15.70) witha 1 (x)=1/x,
a 0 (x)=(x^2 −1)/ 4 x^2 andf(x) = 0. Therefore, making the substitution

y=vu=vexp

(

−

∫

1

2 x

dx

)

=

Av

√

x

,

we obtain

d^2 v

dx^2

+

v

4

=0. (15.75)

Equation (15.75) is easily solved to give

v=c 1 sin^12 x+c 2 cos^12 x,

so the solution of (15.74) is

y=

v

√

x

=

c 1 sin^12 x+c 2 cos^12 x

√

x

.

As an alternative to choosingu(x) such that the coefficient ofv′in (15.71) is

zero, we could choose a differentu(x) such that the coefficient ofvvanishes. For

this to be the case, we see from (15.71) that we would require

u′′+a 1 u′+a 0 u=0,

sou(x) would have to be a solution of the original ODE with the RHS set to

zero, i.e. part of the complementary function. If such a solution were known then

the substitutiony=uvwould yield an equation with no term inv, which could

be solved by two straightforward integrations. This is a special (second-order)

case of the method discussed in subsection 15.2.3.

Solution method.Write the equation in the form (15.70), then substitutey=uv,

whereu(x)is given by (15.72). This leads to an equation of the form (15.73), in

which there is no term indv/dxand which may be easier to solve. Alternatively,

if part of the complementary function is known then follow the method of subsec-

tion 15.2.3.