Name
SotvnntJc
Student ID.
______________
Math 2250-4
Exam 2
March 28, 2013
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. Good Luck!
Score
POSSIBLE
v-z~
VI
1’
1
la) Are
2
-2
3
1
2
25
3
25
4
20
5
10
100
II
1
II
,
20
TOTAL
-J
v2,
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I
9
,
1
a basis for R~ ? If they are, explain why. If they aren’t a basis for
~,
1
determine what sort of subspace of R3 they do span, and fmd a basis for that subspace.
(10 points)
r’
‘1
9
-z
~2
3
I
I
ta~a≤4~1
4-Lt~
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-
rr~(A~t
So
tc ,f\zH4, -J
I~(~-t- j/~l~,
lifLi
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~
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~tQ
-~
a rr4
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(&~≤c~p~)
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~A\~
tQ
A~ ~I
I
I b) Arey1 (x)
=
e2 ~ y2 (x)
=
I a basis for the solution space of the differential equation
y’’’(x) —2y’’(x)O?
If so, explain
the solution space.
(10 points)
s2L
Sc, bo~~cs fr~.uToLc 2
(3
r —Zr
(ii
kA~’s~
r (r~).
2%
Sp4~cL
2) 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
mx’’ + cx’ + kx=F0 cos(cot)
with m,k,o> 0; c,F0 ≥ 0.
2a) Explain what each of the letters m, As c, F0, w represent in this model. Also give their units in the mk~
system.
(10 points)
(~a/~~)
F0
fD~ ~
2~ L~- Ftc~R,
&tI,JS
ccorc~~ft
&fr%~Ja ef~rt1va~L~
(r~j4)
Explain what values of c, F0, w lead to the phenomena listed below. 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.)
( vrf,{~a-%Q
2b)_underdamped oscillations
~ L~ 0
L~c 2 ~ç~x ~h ~
X”t C~’
C,
1’
i—
I
%.t,(
j
t(
cLtcott~,O≥4 oO’I~
(~1 points)tScilt44~
.1
c-QSvJL
2d) beating
~èco
kw~)L
t
F0L4cL,) 4
(4 points)
2e) practical resonance.
(4 points)
/t~c~/*Lxt
L~S~t
3a) Use Chapter 5 techniques to solve the initial value problem
x’’(t) + 6x’(t) + 9x(1) = l8cos(3 t)
x(0)0
x’(O) = 2
(20 points)
Hint: first use the method of undetermined coefficients to find a particular solution. (This is the ‘easy’
Case 1 of undetermined coefficients, because there is damping!) Even if you can’t find the correct
undetermined coefficients, you can still proceed through the rest of the problem to find the IVP solution, in
terms of those undetermined coefficients. You will get partial credit for doing so.
-z
+-
~
—;A~h~t ~3S&os~L
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i-iç
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(L~)
l_
~ ≤i~3t
~r~)
tjq~to
~
* ≤i~tt
(o)
t’~P
A-zo
x(o~HO~
O+-~i
-I-
x’(o~ 2
‘r~7
a
‘cV ptr’)
r2÷~6rij
~-t1~ c,1e
-~
~2*=
~ CI ~
~z(rfs)
i
-‘~
~
-
(3-c’,
sf~t3t *-CL
-~
j
~2
3b) Write the steady periodic part of your solution to part (a) in amplitude phase form,
x(t) = Ccos(3 r— a).
6
1~c~o~3b
Co ,t
(5 points)
‘~oc)
,~ tt)
-~
C
L05
÷Bs1’~,tr
(3~
cC~
4) Use the Laplace transform technique to re-solve the same IVP as in problem (3):
x’’(t) + 6x’(t) + 9x(t) = l8cos(3 t)
x(0) = 0
x’(O) =2
If you cant find the correct partial fraction 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.
(20 points)
G
g
X(~) (≤~Gs~-’1)
~ 9~(c)~
S
≤
1-~
t
Ac#B
4-2
2-
C
S4-3
2
St-I
•~
1)
~g t2
—
/k
hA
Os3
i-Qçt
:~ j (~B~29C k
—
3
—
4~2~ti
—
3
2
L~~)
(~B)
— •2-
i-~, fIC
+2~
~~st(÷3~
÷I~s
~c
~stt9)
2-
43C
1€
~
5) Use the Laplace transform table to find the inverse Laplace transform of
4s+1O
+
(s2+2s+10)
(s2+4)2
(10 points)
4(s~ i4,
E(s~
F~
_________
—
I
£4’
+
(s4o 4-3
4 ~ ~
~ ~ -2~ws2f)]
Table of Laplace flansfonus
This table summarizes the general properties of Laplace transforms and the Laplace transforms of particular functions
derived in chapter 10.
~
f(s)
F(s)
af(t) + bgQ)
oF(s) + bG(s)
f’Q)
sF(s)
f”(g)
s2F(s)
—
sf(0)
f°°(t)
s’F(s)
—
s’’f(O)
—
f(0)
—
f’(O)
—
F(s)
f
—
f°’~(O)
I~teW
(s
coskr
s2+k2
sinks
c2+k2
coshkt
k
.
—
sinhkt
e4f(t)
F(s —a)
e~’ eoskt
uQ—a)f(t—a)
r”~F(s)
esink:
j f(r)gQ
F(s)G(s)
~~(sinkr
f(r)dr
—
r) dr
if (,)
I”
f(t)
(s—a)2
—
ki cosk;)
—F’(s)
sin kg
(—I )°F’° ts)
(sin ki ÷ ki cos kg)
2
I
f(s). period p
u(t—a)
s_os
—
5
I
f”c’fwds
ô(r —a)
I.
(—l)~”°~
—
II — i
lj_aJj
till
4
so
(~2 +k2)2
-
F(c)dcrc
j
Js
+ k’)’
(s2 ± k2)’
fOZ
—
—~~‘
r(a + I)
5(1+1
(square wave)
.
(staircase)
‘lanh
______
s(I
—
rus)