Numerical Analysis in Engineering
ME 140A, Fall 2008
Midterm #2 Solution
By: Mohamad M. NasrAzadani
[email protected]
December 3, 2008
Problem 1
Using separation of variables, the given ODE can be rearranged as
y
2
dy
=
2
dx
x
.
(1)
Integrating both sides of equation (1) will give us
Z
y
2
dy
=
Z
2
dx
x
y
3
3
=
2 ln

x

+
C
(2)
Here,
C
is a constant which can be found using the initial condition. Equation
(2) gives
y
as a function of
x
implicitly. Finally,
y
is found as
y
= (6 ln

x

+ 3
C
)
1
3
.
(3)
1
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For the ﬁrst order ODE with the form given by
y
0
+
p
(
x
) =
g
(
x
)
(4)
the integrating factor
μ
(
x
) is found by
μ
(
x
) =
e
R
p
(
x
)
dx
.
(5)
Hence, we have
μ
(
x
) =
e
R
2
xdx
=
e
x
2
(6)
Multiplying
μ
(
x
) to the both sides of equation (4), we have
d
dx
[
μy
] =
g
(
x
)
μ
(
x
)
(7)
Integrating both sides of equation (7) results in
μ
(
x
)
y
=
Z
g
(
x
)
μ
(
x
)
dx
(8)
Using equation (6), we have
e
x
2
y
=
Z
xe
x
2
dx
=
Z
1
2
(2
x
)
e
x
2
dx
=
1
2
e
x
2
+
C
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 Spring '08
 Meiburg
 Quadratic equation, Elementary algebra, particular solution yp

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