8.11.2. Example

Find the general solution of

        y y= 2 1+ e x

on the interval ( , ) .

Solution

The independent solutions of the homogenous equation ( D 2 1 )y=0  are

        u 1 ( x )= e x           and        u 2 ( x )= e x

The Wronkian matrix of u1 and u2 is

        W( x )=( e x e x e x e x )

with inverse

        W 1 ( x )= 1 2 ( e x e x e x e x )

Hence,

        W 1 ( x ) E n = W 1 ( x )( 0 1 )= 1 2 ( e x e x )

        R( t ) W 1 ( t ) E n = 1 1+ e t ( e t e t )

so that

        ν 1 ( x )= 0 x dt    e t 1+ e t   = e x x+ln( 1+ e x )+1ln2

        ν 2 ( x )= 0 x dt    e t 1+ e t   =ln( 1+ e x )+ln2

Since the constant terms can be absorbed into the homogeneous solutions, we have

        y 1 =[ e x x+ln( 1+ e x ) ] u 1 +[ ln( 1+ e x ) ] u 2

                =1x e x +( e x e x )ln( 1+ e x )

so that the general solution is

        y=1x e x +( e x e x )ln( 1+ e x )+ c 1 e x + c 2 e x