7)
Name.
I.D. number
.
Math 2280-001
Second Midterm Exam
April 3,2015
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, and the point values for each problem are indicated in the right-hand margin.
Good Luck!
Score
possible
1
20
2
25
3
30
4
5
5
20
total
100
fl
Consider the following initial value problem, which could arise from Newton’s second law in a forced
mass-spring oscillation problem:
x’’ (t) + 2x’ (t) + lOx(t) = 26 sin(2 t)
x(O)0
x’(0)=6.
j~ As a step in solving this IVP, fmd the general solution to the (underdamped) homogenous differential
equation
x’’(t) +2x’(t) + lOx(t) =0
(5 pomts)
~t+)c C,e~ç~
ft).
&c~
Find a particular solution to
x’’(t) +2x’(t) + lOx(t)26sin(2t)
using the method of undetermined coefficients.
± i~
-~-
(
%(
‘
~
~
C
~
~~-z
(10 points)
I
~o~if ~
K1,—
~1A~ocz{
i~c~i&)
LC~~ ~s~f( \oA+~B4’’’
~≤r~f(
LoS
(2-k) (—9A k-≤-13)
2.~(oSCv~2~t
tArn’~ 0
4— St’v
t7~StW\4
Ic)
-=
r3 -ziro
L2~ ~1L’~i
3,A~f2i~
K1,t —2~s2f
~ ≤1~4 E
Use your work from parts a,b to solve the initial value problem at the top of this page.
(5 points)
‘4’
~2~szt43si\~2-t
*c2e
i-c~c
S IVY)
X(o’y’—O-t—Z4-c~ -4c~~2
C
c1t
3c2~
-)
0
t
c2~= ~
2
Z~)
The homogeneous differential equation that you solved in k namely
x’’(t) +2x’(t) + lOx(t) =0
corresponds to the following f~1~rder system of two differential equations:
r~r~
[x~’(t)
._<...~
~
0
=[ -10
[~I~
1 1[Xi
-2
1
Careflully explain the correspondence between solutions x(t) to the second order differential equation and
[xi(t)
solutions
L4~
xafl
I
1
1to the first order system.
[x2(t)]
~.
~
LZ~]
(5 pomts)
‘cs
Se~S®)
)S/~(4’~’t_b0a hc’
Q
~ Use your work from jg) to write down the general (homogenous) solution to the first order system
F
x
~ ‘1
[x2’(t)
~
p[
0
1
i[x1
-10 -2 i[x2
1
j
Hint: It would be ugly to solve this problem using the eigenvalue-eigenveetor methods of Chapter 5.
(5 points)
~ 1
L~”~’i
dt~~
~
c~eE t~tc3.b
-t
≤~-3b
;~ts~3~)
i—cc
c1(_&t~53~
—i; ~53~
t~53~ 4-~s”~~;t
~c
~c~c
(—e~~s~n3t
tr
+;e~s34)
~f~~3t
+to~3L
3
~g) Find the (complex) eigenvalues for the matrix in ZJz. Use this information to classify the phase portrait
as one of: nodal source, nodal sink, saddle, spiral source, spiral sink, center. For your convenience, the
system is repeated here:
x1’(t)
—
x21(t)
—
0
1
X1
-10 -2
(5 points)
+ iC
-
—10
—za
(~i)2t~
@41Nt
to
-1
)<o.
~~t3C
~rJsiSL
~s~)
Sketch the phase diagram for the system in ~ Hint: It is possible to construct an adequate picture by
using the classification in ~ and by checking the tangent vector field in several places.
(10 points)
1
[~1
~
-tj
L~ j
.3
ro
ci
L~1
Cr
—3
cc ~
V’oi.
4
fl
Consider a general input-output model with two compartments as indicated below. The compartments
contain volumes V~, 1’~ and solute amounts x1 (t), x2 (t) respectively. The flow rates (volume per time) are
indicated by r. I = 1 ..6. The two input concentrations (solute amount per volume) are c1, c5.
,
3~ Suppose ~2
=
73
=
=
100, r1
=
74
=
200,75
=
0
.
Explain why the volumes V~ (t), V2(t)
remain constant.
c
V7~
~
—ç—r~
2~ ~
(4points)
y
≤~
V~ o°
V1~
3b~ Usiiw the flow rates above. c I
—
=
-
0.6. c
0 gaL’
V
=
-~—.
=
V2
=
100 gal. show that the amounts of
-
solutex1(t) intanlc I andx2(t) intank2satisfS’
—3
x1’(t)
x2(t)
X1-r~c1 ~_r~
ç~ .i-r3
~5-.
26b (4)- 1&v £i
-V2~O
39
2 —2
x2
120
+
0
(6 points)
—~
1-LdD
~.
‘~t
—Z5DI.X
~
1,’t3K4L1c)
)(‘t~
=
1
Y;ct ~
\
~j
—200
~
5
Find the general solution to the system
x1’(t)
x2’(t)
—
—
_3
2
I
—2
][
xI
120
+
0’
Hint: As steps you will want to find the homogeneous solution
and then a parti lar solution which is
a constant vector, x1, = C.
_2
~ H’~
F3
~ J L~
‘1çx~
2JL’S-
t\3 499~Lt
C
4’-)
—2c1 4 ~Z-O -c~
—‘-GO
It,0
L-~J
-J
0
~4€i
V
Solve the initial value problem for ~ assuming there is initially
4-
[
-il
C-
solute in either tank.
(5 points)
—‘It
LGOJ
La] L~J
b
-=
~
e_t[[]
r~-o~
•~) <~c
~~lCL
-
10
—t U
‘2~
~=o.
e-2c1
points)
C~1~[’~
jt[~
a
0
~)
(15
_j
+
-I- ç
U-
—I
~u~t
fC~[l~
tcz[~j
—2-0
I
_~~~t[1]_9o~
ét ~—9Oe~
4-2-OC
r2
31
-
L’
—,
-
14~ LI-ir-~
-31
H2-oJ
L~1b
6
41
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(on)
withm,/çw> 0; c,F0 ≥ 0.
When the forcing amplitude F0 is positive there are two related but not identical phenomena known as
“pure resonance” and “practical resonance” that can arise. Explain what each of these is, how they arise (in
terms of the the parameters m, k, w, c), how they are similar and how they are different.
(5 points)
rLSdkeCt
C~O
~
Lc
~
hQp~1~
~$~7~V’~
s~L~
)ç~
( t
~
4
(~ ~5~± 4çsi~~
cpldkfl ç~rS3
C.)O
~‘~flL
n~4-
tC>c’tLKtL
s~Li,
F ~
~
“-X~
~~€) ~
~
C
I’
car-..
-f~t~g~
~ç~LAi~At
C~CM
~a~-
~
tt~oc(k4s4,.
I
~2tcAtq
~=) K 15
c’,c
Ci
(~V2~4S
&o~4~’o~~
Il~
~€~-
~
s~
rac-{,tJ2-
v-es 6k~4thCO—~
7
~
Ct
a Consider the following undamped mass-spring configuration, where as usual x1 (t), x2 (1) measure the
displacement of the two masses from equilibrium.
Ic
~j In case in1
=
2, m2
=
1, ~1
=
4, /c~
=
2,/c3
=
0 (in other words, the third spring isn’t really there), show
that the resulting system of differential equations reduces to
x1 ‘‘(t) =—3x1 +
(
W~\ )(~
x2”(t)=2x1—2x2
—(L14-1~)x~, .c-L<2~,~2~
(5points)
4h2jC2”tz 1<2x1
(G’)c ~2x,
—(k2#L<~)~2
What is the dimension of the solution space to this system of DEs? Explain.
(3 points)
e~
¾
cctle~
÷
V23 C
bE’~s
nJwwoa)
(cisc
~kt%tcjl £S~-o($ &4,~1~-~
G,..L cS t~Qt4o)J)
~ Find the generaY solution to this system of differential eq~thtions. (Hint: The matrix in this second
order system is the same as the matrix of the first order system in 3, and you may use that eigendata
without rederiving it).
(8 points)
r~’i~H
‘1r~i
[c”J Li~
-2JLx~J
I
(c1wsat
p
t
j,2.
Describe the two fundamental modes of this system.
(4 points)
rl,
2~ kAs
.{w.a
pL1~4tt~ t1~.A5c
4.
~4c~tth2~l
~“ (
a
t1~
8