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Exercises 4.5
37.
Clearly,
r
= 0 is not a root of the auxiliary equation,
r
3
−
2
r
2
−
r
+ 2 = 0. (One can Fnd
the roots, say, using the factorization
r
3
−
2
r
2
−
r
+2=(
r
−
2)(
r
−
1)(
r
+ 1), but they are
not needed for the form of a particular solution: the only important thing is that they are
di±erent from zero.) Therefore, a particular solution has the form
y
p
(
t
)=
A
2
t
2
+
A
1
t
+
A
0
.
Substitution into the original equation yields
y
0
p
−
2
y
0
p
−
y
0
p
+2
y
p
=(
0
)
−
2(2
A
2
)
−
(2
A
2
t
+
A
1
)+2(
A
2
t
2
+
A
1
t
+
A
0
)
=2
A
2
t
2
+(
A
1
−
2
A
2
)
t
+(
A
0
−
A
1
−
4
A
2
)=2
t
2
+4
t
−
9
.
Equating the coeﬃcients, we obtain
2
A
2
=2
,
2
A
1
−
2
A
2
=4
,
2
A
0
−
A
1
−
4
A
2
=
−
9
⇒
A
2
=1
,
A
1
=3
,
A
0
=
−
1
.
Therefore,
y
p
(
t
)=
t
2
+3
t
−
1.
39.
The auxiliary equation in this problem is
r
3
+
r
2
−
2 = 0. By inspection, we see that
r
=0is
not a root. Next, we Fnd that
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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