Chapter 4
y
p
(
θ
)
=
2 (
A
cos
θ
+
B
sin
θ
)
e
−
θ
+
θ
(
A
cos
θ
+
B
sin
θ
)
e
−
θ
=
2 [(
B
−
A
) cos
θ
−
(
B
+
A
) sin
θ
]
e
−
θ
+
θ
(
A
cos
θ
+
B
sin
θ
)
e
−
θ
.
(Note that we did not evaluate the terms containing the factor
θ
because they give zero result
when substituted into the original equation.) Therefore,
y
p
+ 2
y
p
+ 2
y
p
=
2 [(
B
−
A
) cos
θ
−
(
B
+
A
) sin
θ
]
e
−
θ
+ 2(
A
cos
θ
+
B
sin
θ
)
e
−
θ
=
2 (
B
cos
θ
−
A
sin
θ
)
e
−
θ
=
e
−
θ
cos
θ .
Hence
A
= 0,
B
= 1
/
2,
y
p
(
θ
) = (1
/
2)
θe
−
θ
sin
θ
, and a general solution is given by
y
(
θ
) =
1
2
θe
−
θ
sin
θ
+ (
c
1
cos
θ
+
c
2
sin
θ
)
e
−
θ
.
23.
The corresponding homogeneous equation,
y
−
y
= 0, is separable. Solving yields
dy
dt
=
y
⇒
dy
y
=
dt
⇒
ln

y

=
t
+
c
⇒
y
=
±
e
c
e
t
=
Ce
t
,
where
C
= 0 is an arbitrary constant.
By inspection,
y
≡
0 is also a solution.
Therefore,
y
h
(
t
) =
Ce
t
, where
C
is an arbitrary constant, is a general solution to the homogeneous
equation. (Alternatively, one can apply the method of solving first order linear equations in
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Sin, Cos, Trigraph, Elementary algebra, corresponding homogeneous equation

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