### 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 ±2*i*. 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 *c*_{k}’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
)=24a−16(
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
(
24a−1
)
=−
1
16
(
5
2
)

so that the desired particular solution is

y
1
=−
1
16
(
x
4
+x+
5
2
)