Chapter 2
At the initial point,
x
=
π/
4, cos(
π/
4)
>
0 and, therefore, we can take
µ
(
x
)=(co
s
x
)
−
1
.
Multiplying the standard form of the given equation by
µ
(
x
)g
ives
1
cos
x
dy
dx
+
sin
x
cos
2
x
y
=2
x
⇒
d
dx
±
1
cos
x
y
²
=2
x
⇒
1
cos
x
y
=
Z
2
xdx
=
x
2
+
C
⇒
y
=cos
x
(
x
2
+
C
)
.
From the initial condition, we ±nd
C
:
−
15
√
2
π
2
32
=
y
³
π
4
´
=cos
π
4
µ
³
π
4
´
2
+
C
¶
⇒
C
=
−
π
2
.
Hence, the solution is given by
y
=cos
x
(
x
2
−
π
2
).
23.
We proceed similarly to Example 2 on page 52 and obtain an analog of the initial value
problem (13), that is,
dy
dt
+5
y
=40
e
−
20
t
,y
(0) = 10
.
(2.10)
Thus
P
(
t
)
≡
5and
µ
(
t
)=exp
(R
5
dt
)
=
e
5
t
. Multiplying the di²erential equation in (2.10)
by
µ
(
t
) and integrating, we obtain
e
5
t
dy
dt
+5
e
5
t
y
=40
e
−
20
t
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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