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Exercises 4.5
a particular solution has the form
y
p,
1
(
x
)=
A
cos
θ
+
B
sin
θ
.
Substitution into the
equation yields
(
−
A
cos
θ
−
B
sin
θ
)
−
(
A
cos
θ
+
B
sin
θ
)=
−
2
A
cos
θ
−
2
B
sin
θ
=s
in
θ
⇒
−
2
A
=0
,
−
2
B
=1
⇒
A
=0
,
B
=
−
1
/
2
.
So,
y
p,
1
(
θ
)=
−
(1
/
2) sin
θ
.
(ii)
For the equation
y
0
−
y
=
e
2
θ
,
a particular solution has the form
y
p,
2
(
θ
)=
Ae
2
θ
. Substitution yields
(
Ae
2
θ
)
0
−
(
Ae
2
θ
)
=3
Ae
2
θ
=
e
2
θ
⇒
A
=1
/
3
,
and
y
p,
2
(
θ
)=(1
/
3)
e
2
θ
.
By the superposition principle, a particular solution to the original nonhomogeneous equation
is given by
y
p
(
θ
)=
y
p,
1
(
θ
)
−
y
p,
2
(
θ
)=
−
(1
/
2) sin
θ
−
(1
/
3)
e
2
θ
,
and a general solution is
y
(
θ
)=
y
p
(
θ
)+
y
h
(
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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