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Chapter 2
19.
For a general solution, separate variables and integrate:
dy
dθ
=
y
sin
θ
⇒
dy
y
=s
in
θdθ
⇒
Z
dy
y
=
Z
sin
θdθ
⇒
ln

y

=
−
cos
θ
+
C
1
⇒
y

=
e
−
cos
θ
+
C
1
=
Ce
−
cos
θ
⇒
y
=
−
Ce
−
cos
θ
(because at the initial point,
θ
=
π
,
y
(
π
)
<
0). We substitute now the initial condition,
y
(
π
)=
−
3, and obtain
−
3=
y
(
π
)=
−
Ce
−
cos
π
=
−
Ce
⇒
C
=3
e
−
1
.
Hence, the answer is given by
y
=
−
3
e
−
1
e
−
cos
θ
=
−
3
e
−
1
−
cos
θ
.
21.
Separate variables to obtain
1
2
(
y
+1)
−
1
/
2
dy
=cos
xdx.
Integrating, we have
(
y
+1)
1
/
2
=s
in
x
+
C.
Using the fact that
y
(
π
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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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