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Math 251
Spring 2004
Midterm Exam I
1.
Both
f
(
t, y
) =
1
y

3
t
2
and
∂f
∂y
=

1
(
y

3
t
2
)
2
have to be continuous. The only condition
needed is that
y
0

3
t
2
0
6
= 0
2.
a) The equilibrium solutions are
y
=

3
,
0
,
1.
b)
y
=

3 is unstable,
y
= 0 is (asymptotically) stable, and
y
= 1 is unstable.
c) lim
t
→∞
y
(
t
) =
∞
d) lim
t
→∞
y
(
t
) = 0
3.
a)
M
(
x, y
) =
y
2
e
xy
2
+ 4
x
3
,
N
(
x, y
) = 2
xye
xy
2
+ 2
∂M
∂y
= 2
xy
3
e
xy
2
+ 2
ye
xy
2
=
∂N
∂x
So the equation is an exact equation.
b)
e
xy
2
+
x
4
+ 2
y
= 5
4.
a) The roots of the characteristic equation
r
2

4
r

5 = 0 are
r
=

1
,
5. Therefore,
the functions
y
1
=
e

t
and
y
2
=
e
5
t
form a pair of fundamental solutions for this second order
linear diFerential equation.
b) General solution is
y
(
t
) =
C
1
y
1
+
C
2
y
2
=
C
1
e

t
+
C
2
e
5
t
.
c)
y
(
t
) =
7
6
e

t
+
5
6
e
5
t
5.
a) (Solved as a separable equation.)
e
y
= sin
t
+
t
+
e
3
(implicit solution), or
y
= ln(sin
t
+
t
+
e
3
)
(explicit solution).
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 Spring '08
 CHEZHONGYUAN
 Math, Differential Equations, Equations, Partial Differential Equations

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