Exercises 2.4
15.
This differential equation is expressed in the variables
r
and
θ
. Since the variables
x
and
y
are dummy variables, this equation is solved in exactly the same way as an equation in
x
and
y
. We will look for a solution with independent variable
θ
and dependent variable
r
. We see
that the differential equation is expressed in the differential form
M
(
r, θ
)
dr
+
N
(
r, θ
)
dθ
= 0
,
where
M
(
r, θ
) = cos
θ
and
N
(
r, θ
) =
−
r
sin
θ
+
e
θ
.
This implies that
M
θ
(
r, θ
) =
−
sin
θ
=
N
r
(
r, θ
)
,
and so the equation is exact.
Therefore, to solve the equation we need to find a function
F
(
r, θ
) that has cos
θ dr
+ (
−
r
sin
θ
+
e
θ
)
dθ
as its total differential. Integrating
M
(
r, θ
) with
respect to
r
we see that
F
(
r, θ
) =
cos
θ dr
=
r
cos
θ
+
g
(
θ
)
⇒
F
θ
(
r, θ
) =
−
r
sin
θ
+
g
(
θ
) =
N
(
r, θ
) =
−
r
sin
θ
+
e
θ
.
Thus we have that
g
(
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, Expression, How to Solve It

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