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Review Problems
we get the following system for
v
0
1
and
v
0
2
(see (9) on page 195 in the text):
v
0
1
(
θ
)cos4
θ
+
v
0
2
(
θ
)sin4
θ
=0
−
4
v
0
1
(
θ
θ
+4
v
0
2
(
θ
θ
=tan4
θ.
Multiplying the Frst equation by sin 4
θ
and the second equation by (1
/
4) cos 4
θ
, and adding
the resulting equations together, we get
v
0
2
(
θ
)=
1
4
sin 4
θ
⇒
v
2
=
−
1
16
cos 4
θ
+
c
3
.
±rom the Frst equation in the above system we also obtain
v
0
1
(
θ
−
v
0
2
(
θ
)tan4
θ
=
−
1
4
sin
2
4
θ
cos 4
θ
=
−
1
4
(sec 4
θ
−
cos 4
θ
)
⇒
v
1
(
θ
−
1
4
Z
(sec 4
θ
−
cos 4
θ
)
dθ
=
−
1
16
ln

sec 4
θ
+tan4
θ

+
1
16
sin 4
θ
+
c
4
.
Taking
c
3
=
c
4
=0,weobtain
y
p
(
θ
±
−
1
16
ln

sec 4
θ
θ

+
1
16
sin 4
θ
²
cos 4
θ
+
±
−
1
16
cos
θ
²
sin 4
θ
=
−
1
16
(cos 4
θ
)ln

sec 4
θ
θ

,
and a general solution to the original equation is
y
(
θ
c
1
cos 4
θ
+
c
2
sin 4
θ
−
1
16
(cos 4
θ

sec 4
θ
θ

.
25.
Since the auxiliary equation, 4
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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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