Name
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Student I.D.
Math 2250-0 10
Exam 2
March28 2014
Please show all work for full credit. This exam is closed book and closed note. You may use a
scientific calculator, but not one which is capable of graphing or of solving differential or linear algebra
equations. In order to receive full or partial credit on any problem, you must show all of your
work and justify your conclusions. There are 100 points possible. The point values for each problem are
indicated in the right-hand margin. There is a Laplace transform table at the end of this test. Good Luck!
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2.).
Consider the linear homogeneous differential equation fory(x)
y”(x) +y’(x) —6y(x)=O.
Find a basis of solution functions for this homogeneous differential equation. Hint: use the characteristic
polynomial.
(8 points)
(43)()
r—3,z
SO
21).
Use your work from 2a and the method of undetermined coefficients to first fmd particular solutions,
and then the general solutiony(x), to
y”(x)
+y’(x)
6y(x)
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4
In part 2b you used the cact that the general solution to any inhomogeneous linear lifferential equation
L (y) =f is the sum of any single particular solution with the general solution to the homogeneous
differential equation, y =
Explain why this fact is true. Hint: Use the “linearity” properties
+
L(
1 +Y
)
7
=
L (y
) + L (Y
1
)
7
L(cy) =cL(y).
(5 points)
so
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cj-4.
j A focus in this course is a careful analysis of the mathematics and physical phenomena exhibited in
forced and unforced mechanical (or electrical) oscillation problems. Using the mass-spring model, we’ve
studied the differential equation for functions x(t) solving
0 cos(0t)
+ cx’ + kx=F
mx’’
with in, k w> 0; c, F
0
0.
Explain what values (or ranges of values) for c, k, F
, w lead to the phenomena listed below. (If you use”
0
in this discussion make sure to explain what it is as well.) What form will the key parts of the
solutions x(t) have in those cases, in order that the physical phenomena be present? (We’re not expecting
the precise formulas for these parts of the solutions, just what their forms will be.)
3.. simple harmonic motion
II
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4-
x R’
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ks
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(2 points)
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1 cft
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t-’-”
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(3 points)
F,
4
L(L
YAfl’
.
beating
(3 points)
&-Jl
st
+-L2çQ
j.). practical resonance.
tf)
cO-4
(-a
(4)
4)
t
2..
S
iflLel,
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(2 points)
6L I)
rpLu Ilc.t
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tc
h- C-
.
0
F
4.. Use Chapter 5 techniques to solve the initial value problem below, which could represent a forced
oscillator problem:
x’’(t) +4x(t) lOcos(3t)
x(O) = I
x’(O) =2
(15 points)
Hint: first use the method of undetermined coefficients to find a particular solution.
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4i2). What is the period of the solution function in 4a? (Note, the solution is a sum of periodic functions,
and they have a common period.)
(5 points)
1’,
b
Tl---
r
oit s2-
.Th
Use the Laplace transform technique to re-solve the same IVP as ii problem (5):
x’’ (t) + 4x(t) 10 cos(3 t)
x(0) = I
x’ (0) = 2
If you cant fmd the correct partial fl-action coefficients for X(s), you can still use the Laplace transform
table to deduce what the solution x(t) is, in terms of these unknown coefficients. You will get partial
credit for doing so.
(15 points)
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Ll) 4
Recall the integral definition of Laplace transform. namely that for any functionf(t), with
, the Laplace transform F(s) £{f(t) } (s) is computed via the integral
1t
If(t) I Ce’
F(s)
=1
f(t)edt, fors
>
M.
Pick ONE of the two parts below to complete. If you try both, indicate clearly which problem you would
like to have graded.
(10 points)
Use the integral definition above to compute the Laplace transform of
f(t) =te 71
(Of course, you may check your answer with the table at the end of this exam.) Hint: use integration by
parts.
b
t
F (c
00
c20
(2-4 1;
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e.
0-0
—-
12—<
j
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1 The reason that Laplace transform is so effective for linear differential equation initial valu
is because of how it transforms derivatives of functions. This is probably why the transformation rules for
how first, second, and nthderivatives of functions transform are right at the start of the table at the end of
this exam. These derivative transformation rules all follow from the one for how first derivatives
transform, namely:
£{f’(t)}(s)sF(s) —f(0).
Use the integral definition of Laplace transform to derive this table entry.
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et
ab
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vf()
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VcUL
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r
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1-
Jo
--
E
0
Table of Laplace Transforms
of particular functions
This table summarizes the general properties of Laplace transforms and the Laplace transforms
derived in Chapter 10,
ansform
Function
Fnctpn Traform
f(t)
F(s)
s—a
af(t)±bg(t)
aF(s)+bG(s)
(s
f’(t)
sF(s)— f(0)
f”(t)
(s)
s
F
2
—
s”F(s)
— [(0)
sf(O)
—
!-
[f(T)dr
c”
—
sf(0)
coskt
a (t
— a)f(i — a)
f
f(r)g(t
— r)dt
f”(0)
2
coshkt
a-
c” cos ki
e F(s)
e”’ sin ki
F(s)G(s)
-(sinkt
—
s—a
(s
— krcoskt)
sinki
—F’(s)
t”f(t)
s)
(—1)”F”
(
1
(sinkt + kcoskt)
JF(J)dc
u(r—a)
1
peodp
I
fef(t)dt
(t
(_l)L
—
Laj
F(a ± 1)
2
2 +k
a)
2
(s
2 +k
(s
)
2
(2
)
2
+k
—a)
11rD
tu
—
2 +k
(s—a)
2
tf(t)
f(t),
2
-i-k
sinhkt
a)
F(s
fQ)
+k
2
s
sin ki
—
—
/
(square wave)
—sI tanh as2
—
e””
(staircase)
s(l
— e””)