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Chapter 2
7.
In this equation,
P
(
x
)
≡−
1and
Q
(
x
)=
e
3
x
. Hence the integrating factor
µ
(
x
)=exp
±Z
P
(
x
)
dx
²
=exp
±Z
(
−
1)
dx
²
=
e
−
x
.
Multiplying both sides of the equation by
µ
(
x
) and integrating, we obtain
e
−
x
dy
dx
−
e
−
x
y
=
e
−
x
e
3
x
=
e
2
x
⇒
d
(
e
−
x
y
)
dx
=
e
2
x
⇒
e
−
x
y
=
Z
e
2
x
dx
=
1
2
e
2
x
+
C
⇒
y
=
±
1
2
e
2
x
+
C
²
e
x
=
e
3
x
2
+
Ce
x
.
9.
This is a linear equation with dependent variable
r
and independent variable
θ
.Theme
thod
we will use to solve this equation is exactly the same as the method we use to solve an
equation in the variables
x
and
y
since these variables are just dummy variables. Thus we
have
P
(
θ
)=tan
θ
and
Q
(
θ
)=sec
θ
which are continuous on any interval not containing odd
multiples of
π/
2. We proceed as usual to Fnd the integrating factor
µ
(
θ
). We have
µ
(
θ
)=exp
±Z
tan
θdθ
²
=
e
−
ln

cos
θ

+
C
=
K
·
1

cos
θ

=
K

sec
θ

,
where
K
=
e
C
.
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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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