8.14.3. Example 2

Solve y 5 y +6y=x e x .

Solution

We start by rewriting the equation as

        L( y )=( D 2 5D+6 )y=R=x e x

The homogeneous equation

        L( y )=( D2 )( D3 )y=0

has independent solutions

        u 1 ( x )= e 2x    and     u 2 ( x )= e 3x

For the particular solution y1, we notice that R=x e x  is annihilated by A= ( D1 ) 2 .  Hence, y1 must satisfies

        ( D1 ) 2 ( D2 )( D3 )y=0

whose general solution is

        y=a e x +bx e x +c e 2x +d e 3x

Since L annihliates the last 2 terms, we can write

        y 1 =a e x +bx e x

so that

        D y 1 =( a+b ) e x +bx e x       and        D 2 y 1 =( a+2b ) e x +bx e x

        L( y 1 )=[ a+2b5( a+b )+6a ] e x +( b5b+6b )x e x

                =( 2a3b ) e x +2bx e x   =R=x e x

Þ            2a3b=0   and  2b=1

Hence              b= 1 2        and    a= 3 2 b= 3 4

The general solution is therefore

        y= c 1 e 2x + c 2 e 3x + 3 4 e x + 1 2 x e x