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Exercises 6.3
By diferentiating
y
p
(
x
), we Fnd
y
0
p
(
x
)=
C
1
xe
x
+
C
1
e
x
+
C
3
+2
C
4
x
⇒
y
0
p
(
x
C
1
xe
x
C
1
e
x
C
4
⇒
y
0
p
(
x
C
1
xe
x
+3
C
1
e
x
.
Substituting these expressions into the original diferential equation, we obtain
y
0
p
(
x
)
−
2
y
0
p
(
x
)
−
5
y
0
p
(
x
)+6
y
p
(
x
C
1
xe
x
C
1
e
x
−
2
C
1
xe
x
−
4
C
1
e
x
−
4
C
4
−
5
C
1
xe
x
−
5
C
1
e
x
−
5
C
3
−
10
C
4
x
+6
C
1
xe
x
C
2
C
3
x
C
4
x
2
=
e
x
+
x
2
⇒−
6
C
1
e
x
+(
−
4
C
4
−
5
C
3
C
2
)+(
−
10
C
4
C
3
)
x
C
4
x
2
=
e
x
+
x
2
.
Equating coeﬃcients yields
−
6
C
1
=1
⇒
C
1
=
−
1
6
,
6
C
4
⇒
C
4
=
1
6
,
−
10
C
4
C
3
=0
⇒
C
3
=
10
C
4
6
=
10
36
=
5
18
,
−
4
C
4
−
5
C
3
C
2
⇒
C
2
=
4
C
4
+5
C
3
6
=
4(1
/
6) + 5(5
/
18)
6
=
37
108
.
Thus, a general solution to the nonhomogeneous equation is given by
y
(
x
y
h
(
x
)+
y
p
(
x
c
1
e
x
+
c
2
e
3
x
+
c
3
e
−
2
x
−
1
6
xe
x
+
1
6
x
2
+
5
18
x
+
37
108
.
7.
In Problem 3, a general solution to the associated homogeneous equation was ±ound to be
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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