Math 293 Prelim #2 Solutions
Spring 2000
1
.
The equation
x
dy
dx
+
y
2
=
x
2
y
2
is separable since it can be rewritten as
dy
dx
=
xy
2

y
2
x
=
y
2
(
x

x

1
)
, from which we get
y

2
dy
=
(
x

x

1
)
dx
. Integrating
both sides gives

y

1
=
x
2
2

ln
x
+
C
, hence
y
=
(
ln
x

x
2
2
+
C
)

1
. Plugging in the
initial condition
y(
1
)
=
1 gives
C
=
3
2
, so
y
=
(
ln
x

x
2
2
+
3
2
)

1
.
2
. (a) An integrating factor for the equation
y
0
+
4
y
=
e

x
is
e
4
dx
=
e
4
x
. Multiplying
the equation by this gives
(e
4
x
y)
0
=
e
3
x
. Integrating both sides gives
e
4
x
y
=
e
3
x
3
+
C
.
Solving for
y
, we get
y
=
e

x
3
+
Ce

4
x
as the general solution.
(b) Think of
y
0
+
4
y
=
e

x
as a secondorder equation in which the coefficient of
y
00
happens to be zero. To solve the associated homogeneous equation we look at the
‘quadratic’ equation 0
r
2
+
r
+
4
=
0, with the root
r
= 
4. This gives the general
solution
y
=
Ce

4
x
for the homogeneous equation. Next we find a particular solution
of the nonhomogeneous equation by the method of undetermined coefficients, trying
for a solution of the form
y
=
Ae

x
.
Computing
y
0
= 
Ae

x
and plugging into
the nonhomogeneous equation, we find that
A
must be
1
3
, so we have the particular
solution
y
=
e

x
3
.
Hence the general solution of the nonhomogeneous equation is
y
=
e

x
3
+
Ce

4
x
, the same answer as in part (a).
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 Spring '07
 TERRELL,R
 Math, Differential Equations, Equations, 2J, general solution, dy

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