MATH 251 Fall 2003 Exam 2
November 10, 2003
Solution Key
1. (5 points) A springmass system, subject to an external force of
10 cos(2
t
)
Newtons,
is equipped with a spring with a Hooke’s constant
12
Newtons per meter. For what
mass will the resonance occur?
(a)
2
kg
(b)
3
kg
Correct
(c)
6
kg
(d)
10
kg
Solution:
The frequency of free oscillation
ω
is subject to the following relation:
ω
2
=
spring constant
mass
For resonance the frequency of the external force should be equal to the frequency of
free oscillation.
2. (5 points) Of what form will the particular solution to the following differential equa
tion be? Do not solve the equation.
y
00

4
y
0
+ 4
y
=
e
2
t
+
t
2
e
3
t

sin(2
πt
)
(a)
Ae
2
t
+
Bt
2
e
3
t
+
Cte
3
t
+
De
3
t
+
E
sin(2
πt
) +
F
cos(2
πt
)
(b)
Ate
2
t
+
Bt
2
e
3
t
+
Cte
3
t
+
De
3
t
+
E
sin(2
πt
) +
F
cos(2
πt
)
(c)
At
2
e
2
t
+
Bt
2
e
3
t
+
Cte
3
t
+
De
3
t
+
E
sin(2
πt
) +
F
cos(2
πt
)
Correct
(d)
Ae
2
t
+
Bt
2
e
3
t
+
E
sin(2
πt
) +
F
cos(2
πt
)
Solution:
The characteristic equation, associated with the homogeneous part of
the equation, has root
2
of multiplicity
2
. Therefore the solutiontobe has the
e
2
t
term
multiplied by
t
2
. For the
t
2
e
3
t
part of the right–hand side we need to throw a full
polynomial of degree
2
multiplied by the exponent function, not just
Bt
2
.
3. (5 points) The second order linear equation
y
00

4
y
0
+ 4
y
= 0
is equivalent to
(a)
(
x
0
1
=
x
2
x
0
2
=

4
x
1
+ 4
x
2
Correct
(b)
(
x
0
1
=
x
1
x
0
2
=

4
x
1
+ 4
x
2
(c)
(
x
0
1
=
x
2
x
0
2
= 4
x
1

4
x
2
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(
x
0
1
=
x
1
x
0
2
= 4
x
1

4
x
2
Solution:
We think of
x
1
as
y
and
x
2
as
y
0
.
4. (5 points) Which of the following functions corresponds to this graph:

t
6
y
¡
¡
¡
¡
¡
¡
¡
¡
¡
¡
0
1
2

1
(a)
tu
1
(
t
)

1
(b)
tu
1
(
t

1)

1
(c)
(
t

1)
u
1
(
t
)

1
Correct
(d)
(
t

1)
u
1
(
t

1)

1
Solution:
We are ’turning’ at
t
= 1
, that’s why we must have
u
1
(
t
)
term. Before
the turnpoint, we are at the level

1
, that’s why we have to add constant

1
. After
the turnpoint, our function looks like
t

2
, therefore we write
t

1
in front of the
Heaviside function (remember, we already have

1
before the turn).
5. (14 points) Solve the following initial value problem:
y
00

5
y
0

14
y
=

14
t
2

10
t

26
,
y
(0) = 0
,
y
0
(0) = 13
Solution:
The associated homogeneous differential equation
y
00

5
y
0

14
y
= 0
has characteristic polynomial
2
points
λ
2

5
λ

14 = (
λ

7)(
λ
+ 2)
with roots
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 Spring '08
 CHEZHONGYUAN
 Math, Differential Equations, Equations, Partial Differential Equations, Laplace, Constant of integration

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