Differential Equations test one solutions
Tuesday, April 25, 2006
1
. Given the function
y
= 4+
x
2
sin
s
2
ds
(
a
) Calculate
y
(2) = 4+
2
2
sin
s
2
ds
= 4+0 = 4
(
b
) Calculate
dy
dx
= sin
x
2
2
. Is the function
y
=
x
+
1
x
a solution of
dy
dx
+
y
2
= 3+
x
2
? Yes because:
dy
dx
+
y
2
=
d
dx
x
+
1
x
+
x
+
1
x
2
=
1

1
x
2
+
x
2
+ 2 +
1
x
2
= 3 +
x
2
3
. Determine whether each of the following equations is linear or nonlinear.
(
a
)
dy
dt
=
t

y
2
nonlinear
(
b
)
t
dy
dt
=
e
t
y

cos
t
2
linear
(
c
)
y
dy
dt
= 4
t

y
nonlinear
4
. Which of the following equations is exact?
(
a
)
3
x
2

4
y
dx
+(2
y

4
x
)
dy
= 0
3
x
2

4
y
y
?
= (2
y

4
x
)
x

4 =

4
exact
(
b
)
dy
dx
+
x
2
y
=
x
or (
x
2
y

x
)
dx
+(1)
dy
= 0
x
2
y

x
y
?
= (1)
x
x
2
= 0
not exact
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5
. Given the initial value problem
t
dy
dt
+
3
t

2
y
=
4
t
+ 1
and
y
(3) = 5
determine the largest
t
interval over which a unique solution is guaranteed to exist. Rewrite
the differential equation as
dy
dt
+
3
t
(
t

2)
y
=
4
t
(
t
+ 1)
The coefficients are not continuous at
t
= 0 and
t
= 2.
The solution to the initial value
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 Spring '09
 EWQFQW
 Boundary value problem, dy

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