B U Department of Mathematics
Math 202 Differential Equations
Spring 2005 First Midterm
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1.
Consider the differential equation (
xy

1)
dx
+ (
x
2

xy
)
dy
= 0.
(a)
Show that this is not an exact equation.
(b)
Find an integrating factor
μ
to make it exact.
(c)
Multiply the equation by
μ
on both sides and solve the differential equation.
(d)
State conditions on
x
and
y
required to make your solution valid.
Solution:
Let
M
(
x, y
) =
xy

1 and
N
(
x, y
) =
x
2

xy
. Then
∂
∂y
M
=
x
;
∂
∂x
N
= 2
x

y
. Since
M
y
=
N
x
, the DE is not exact.
We find
μ
(
x, y
) such that
∂
∂y
(
μM
) =
∂
∂x
(
μN
). We require:
μ
y
(
xy

1) +
xμ
=
μ
x
(
x
2

xy
) + (2
x

y
)
μ.
Assuming that
μ
=
μ
(
x
) i.e.
μ
y
= 0 we get:

xμ
x
=
μ
which is in accordance with our assumption. Hence we get as an integrating factor, for
example,
μ
=
1
x
,
x
= 0.
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 Winter '11
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 Equations, Boundary value problem, U Department of Mathematics, Bo˘azi¸i University Mathematics Department

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