### 8.14.2. Example 1

Find a particular solution of

( D 4 16 )y= x 4 +x+1                    (a)

#### Soultion

The right-hand side, being a polynomial of degree 4, can be annihilated by the operator D 5 .  Hence, a solution y of (a) also satisfies

D 5 ( D 4 16 )y=0                            (8.26)

The roots of the associated polynomials are 0, 0, 0, 0, 0, ±2, and ±2i.  Hence,

y= c 1 + c 2 x+ c 3 x 2 + c 4 x 3 + c 5 x 4 + c 6 e 2x + c 7 e 2x + c 8 cos2x+ c 9 sin2x             (b)

Our task is to find a set of ck’s so that

L( y )=( D 4 16 )y= x 4 +x+1

Since the last 4 terms in (b) are annihilated by L, we can set c 6 = c 7 = c 8 = c 9 =0 Furthermore, to avoid working with subscripts, we set

y 1 =a x 4 +b x 3 +c x 2 +dx+e

so that

D 4 y 1 =4!a=24a

L( y 1 )=24a16( a x 4 +b x 3 +c x 2 +dx+e )   = x 4 +x+1

Since this holds for all x, the coefficient of each power of x must vanish.  Hence,

a= 1 16          b=c=0         d= 1 16            e= 1 16 ( 24a1 ) = 1 16 ( 5 2 )

so that the desired particular solution is

y 1 = 1 16 ( x 4 +x+ 5 2 )