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and a general solution is given by
y
(
t
) =

t
cos
t
2
+
c
1
cos
t
+
c
2
sin
t.
The ﬁrst boundary condition,
y
(0) = 0, yields
c
1
= 0. But this implies that
y
(
π
) =

t
cos
t
2
+
c
2
sin
t
±
±
±
±
t
=
π
=
π
2
6
= 1
,
for any constant
c
2
.
48. (a)
Using the superposition principle (Theorem 3), we conclude that the functions
y
1
(
t
) =
(
t
2
+ 1 +
e
t
cos
t
+
e
t
sin
t
)

(
t
2
+ 1 +
e
t
cos
t
)
=
e
t
sin
t,
y
2
(
t
) =
(
t
2
+ 1 +
e
t
cos
t
+
e
t
sin
t
)

(
t
2
+ 1 +
e
t
sin
t
)
=
e
t
cos
t
are solutions to the corresponding homogeneous equation. These two functions are
linearly independent on (
∞
,
∞
) since neither one is a constant multiple of the
other.
(b)
Substituting, say,
y
1
(
t
) into the corresponding homogeneous equation yields
(
e
t
sin
t
)
00
+
p
(
e
t
sin
t
)
0
+
q
(
e
t
sin
t
)
= (2 +
p
)
e
t
cos
t
+ (
p
+
q
)
e
t
sin
t
= 0
.
Therefore,
p
=

2,
q
=

p
= 2, and so the equation becomes
y
00

2
y
0
+ 2
y
=
g
(
t
)
.
(4.3)
Another way to recover
p
and
q
is to use the results of Section 4.3. The functions
y
1
(
t
) and
y
2
(
t
) ﬁt the form of two linearly independent solutions in the case when
the auxiliary equation has complex roots
α
±
βi
. Here,
α
=
β
= 1. Thus, the
auxiliary equation must be
[
r

(1 +
i
)]
·
[
r

(1

i
)] = (
r

1)
2
+ 1 =
r
2

2
r
+ 2
,
leading to the same conclusion about
p
and
q
.
To ﬁnd
g
(
t
), one can just substitute either of three given functions into (4.3). But
we can simplify computations noting that, say,
y
=
t
2
+ 1 +
e
t
cos
t

y
2
(
t
) =
t
2
+ 1
is a solution to the given equation (by the superposition principle). Thus, we have
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 Spring '08
 MAZMANI
 Equations

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