I.D. number
.
Math 2280-001 Spring 2015
FINAL EXAM
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. A Laplace Transform table and
particular solution table are included with this exam. In order to receive full or partial credit on any
problem, you must show all of your work and justi~’ your conclusions. This exam counts for 30% of
your course grade. It has been written so that there are 150 points possible, and the point values for each
problem are indicated in the right-hand margin. Good Luck!
problem
score
possible
1
25
2
15
3
30
4
25
5
15
6
10
7
15
8
15
total
150
•~ Find the eigenvalues and eigenspace bases for the following matrix:
—3
1
A=
2 -4
Hint: The characteristic polynomial has integer roots.
(8 points)
~t7~
~i2—2
~~+~#(O
H°
210
L-’
It
Check that A j~ = z for each eigenpair (X,v) that you found in part a. This is to catch any mistakes
you may have made, since the matrix in A reappears frequently in this exam.
(2 points)
ft).
ç~3
~ UL’i L~J
~iL—ii
LIbJ
L’i
/
1-u
2
IA
j.g~ Fmd e for the matrix m part a. There are two approaches you may take: either use your work from
part ato first find a fundamental matrix (also known as a non-singular Wronskian matrix) for the system
x’ (t) = A ,~, and work from there; or, useA S = S A (in the form A = S A
to compute the power series
for e~4 directly. If you successfully compute et~~ both ways you will receive 10 extra credit points.
(15 points)
(10 extra credit points also possible)
-J(
-klct
Sb
A-)
X
t/\x
ret
63b
£
[L]
~e
€
t
[et
~—
H
i(~
~ -ze~~J L
IF?
-;
1__~
f~I2tec~
~,
~
~a
-~~iL1 ~‘i
S
I
~lt_t3tt’~
.~tt -c&]
e. t2e
J
~.
2e~2t~
~
a
~
2-
~s$ s_I ~
T4tt~t~-4-- ]s’
SeS
C
-s
~~
L1
~Ar\
e
—~J
~t\
e4~”~’
3Lt
—I
r’L’
iire~-4
~~~Qd_Z_t
(As
.≤cu-.-i
0
-\
11ctJ~
A\~flTh-t)
—‘
—I
3
c~’s ~Lcprt.
~)
Use Laplace transform techniques to find the general solution to the undamped forced oscillator
equation with resonance:
F
2
x’’(t) +w0x(t)= —9--cos(c09t)
x(O)=x0
x’ (0)
ctXtc~
~
—5x0--V~
)KCc)
=
(c’L~%t)
(10 points)
~F0
—j;;;
S
-~
~..
cx0
4-
V0
X(s)~E~ __c
2h)
Consider the more general forced oscillation problem,
x(t) =f(t)
x” (/) +
x(0) =x0
x’(O) =v0.
Find a formula forx(t), that will be valid no matter what functionf(t) is used to force the system (as long
asf(t) is piecewise continuous with at most exponential growth, so that it has a Laplace transform). Hint:
part of your solution will be a convolution integral involving the forcing functionf
g~
)as~)
—~~c _-v
~
(5 points)
~-
4
FV)
4-
___
V0
K05
F
-—
%
~ t+tt~t
4
(9(S)
—
xjt~t
~
4
0
o
4
0
Pet~ t30 ≤1~+~c*~x))Ar ~
0
4
Consider a general input-output model with two compartments as indicated below. The compartments
contain volumes J~, V~ 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.
C
I
X,t~)
3~
What is the system of 4 first order differential equations governing the volumes V1 (t), J/~ (t) and
solute amounts x1 (t), x2 (t)?
V1tit)
V7j
r1
(t.)
(8 points)
—VyY~,
4—’~3!J
—&~+r3~-
~ r~ —Cr ~
X
Suppose r2
-~j-r9
y-v~
)q’~r~c1
3~
tr~
=
r3
100, r~
=
=
Vi
r4 = r5
=
‘~
t
200, r6 = 300 hour
Veri~’ from your work in hthat the
volumes V1 (t), J’~ (t) remain constant.
~
‘çç’
V1
≤~
(2points)
iov÷Z(0~~’3t0~~
3g~Using the flow rates above, c1
=
0.05, c5
=
0.3
V1
=
=
100 gal,show that the amounts of
solutex1(t) in tank 1 and x2(t) intank2satisf~’
x1(t)
3
=
1
X1
+
2 —4
x2’(t)
10
60
(5
</~
!~
~jC1 4-
points)
‘1
2-
.i-Iopxz.
~-(3oo)~~
-3~
a-x~i-u~
IOU
~
v;~,L~
j-
r9
.~
r(rfr&)X2
100
5
2~)
Find the general solution to
x11(t)
x21(t)
—
I
~i
2 —4
x2
_~
—
10
60
+
Note that the matrix in this problem is the same as the one in problem j, You may refer to your results
from that problem. Hint: You may use ~ =
+ ~, Laplace transforms, or any other method we’ve
U]
H
discussed in this course, in order to find the general solution.
c~
c,
r
-Fr
x
3
~-
c~ est
r ~Wt~’=
c~cL
~
(10 points)
I
~
L~ -~JL~j L’°J
~3,ci4ct+u5tz Q
+2Lt~{LL 4-~0zQ
‘4 C1 i-L1 t~—iO
tc~ 2c1~”l0
tE1-i-E_~
_-~z_--’~o
~-L’)
c~LQ’~ c1t-2c?
ID
£t~ct~
2~) Use yo
t:5Fj51
work from d o solve the initial value problem—___.
x1’(t)
—3
1
X1
10
—
x2’(t)
~ -L
—
2
—4
60
x1(O)
—
20
x2(O)
—
0
Ct\ocL~ t ~1C~
(5 points)
C.~4L~
—20
~rE~=r)
~
I
~
‘IH
~X?~b)J
z~—2c~
—5-bil
li-We
L~oJ
6
4)
Although we usually use a mass~spring configuration to give context for studying second order
differential equations, the rigid-rod pendulum also effectively exhibits several key ideas from this course.
Recall that in the undamped version of this configuration, we let the pendulum rod length be L, assume the
rod is massless, and that there is a mass m attached at the end on which the vertical graviational force acts
with force m g. This mass will swing in a circular arc of signed arclength s = L •0 from the vertical, where
0 is the angle in radians from vertical. The configuration is indicated below.
L~Lc.o,o
cCLOj7~
4~) Use the fact that the undamped system is conservative, to derive the differential equation for 0(t),
0’’(t) + fsin(o(t))=o.
(10 points)
Hint: Begin by express the TE=KE+PE in terms of the function 0(t) and its derivatives. Then compute
TE’ (t) and set it equal to zero.
k
2
TE~kL#~’E
-=
V41
Le~u)
L~- L-L~se
(L-L(459Lt’)
k~j
oa
~.q
frLaxef&ht
~
o
t
~-
o~
(at
0
?
7
4j~) For small oscillations (0(t)
linearization
—
0) we replaced the non-linear differential equation in g with the
(C)
0”(t) +
fo(t)=o.
Use the Taylor series for sin ( o) to explain why this is a good approximation to the exact differential
equation in ~, when e.g. el < 0.1 (radians).
(5 points)
~o-~ e—
4
2~#Q4
IsivQ—Sj<
i-.--
L~
w”~ ~
~ora~R14a~%4~
tL~co(t~k
4g)
-~
~
~
+t,sn’.c,c+ 4~.
~j€i ~Ooe2
Carefully describe the connection between solutions 0(t) to the second order linear differential
equation in hand solutions
‘
‘
to the first order system of differential equations
x2(t)
x2
g
x21(1) =—-1--x1.
‘(t)
(i)
=
(4 points)
Gun
s0Q~’ts
-l~l~%a4-_Sc?
U)
Jt’f~i....e
61.4)
X\1t~.~
r~i
[~I
~
Si-U-a
[~]
Ct)
s~≤ (LL~
r-e~
Se, y~ cr?~-tç
(i’).)
8
4d~
In case the numerical value off
=
1 the differential equation in hbecomes
O’’(I) +O(t)=O
and the corresponding first order system in ~ becomes
x11(t)
0 i
xj]
—
x21(t)
—
—1
o
x2j
~
H~
Sketch the phase portrait for the first order system, and classif~’ the origin as one of: nodal source, nodal
sink, saddle point, spiral source, spiral sink, stable center.
(6 points)
(atc)
I
\z
Sc;
I
C
(;~-~)
~
~
s.
V. ~J
41
L~]
‘C-’
f~CLGtCcfl%jZ\~D(
tC~Q svt~c,
@~t)
~u+&zo
5re1a~ThA
~
eu~
L
~Skk~-~.
Cc~oc(&—cc)
r&1 ~
L e j [-c≤t~~j
a~-&
ctve
sc’~-& tL~
‘-~r\~~s&
c4’~~Us
CAQcA&Ar~(
~cA~a,Ltl~~t Ct~tnL5
rc-cLt~ç
C, -h-W,e-$,
a!rm.a-tk 4~t~t
9
a~
Find the general solution to the system of differential equations
x1’’ (1)
—
—3
1
x2’’(t)
—
2
—4
X1
(8 points)
Hint: This second order system of DEs could be modeling a two-mass, three-spring system without
damping and so it will have solutions that oscillate. Also, this is the same matrix as in problem 1 and you
may use results from that problem.
+
1~’N
ci~d~~fl
\~1x2*~ j
~)
Tdentif~’ and describe the
fundamental modes of oscillation in the system a~
(2 points)
;5~t~
W~45
~‘-+LSR~
~a~.pL1~t
—
~
I’wz4c
1.
Ln~dr 2 it
2.
1~.tL
‘~‘+t~
10
~g) Setm1
1, m~
=
-~-,
k~
=
2, k2
=
1, k~
=
1. Show that the displacements x1 (1),x2(t) of the two
masses from equilibrium in the configuration below satis~’ the system in part ~g i.e.
(t) =—3
+
‘‘
x2’’ (t)
=
2 x1
—
4
Hint: Use Newton’s second law that mass times acceleration equals net forces.
(5 points)
hA\)~_
~
-
:~-
~
—çLg,-%’~.— c~,ç
4-
K.
4-
‘,~
11
&We consider a 2 it-periodic tent-wave function, given on the interval (-it, it) byf(t)
1~1•
Here’s a piece of the graph of this function.
-3it--2m
0
-~
~
2m
3m
t
Derive the Fourier series forf(t),
—z
It
2
Fw~
-4cos(nt)
Itnoddfl
to~
b~
(10 points)
—if
5
0~tA.A~
e-~~s--ca
~tic-~,
L~trkVi>L4L.
It
-It’
0~
-fl
so
—
TI—
w
~
ttl
t~
St
Qv-t4--.-It
~ \
L—~
~—__--~
ok-V
Vt
___
I’
~4
•1
S
1°
0
~
I—
U
1.00
~ ~ +Zcs~t
a
-
El
~1-
Tt~’
J~Qdoc½1t_l)
12
fl
Consider the tent-wave funetionf(t) from problem 6, and the forced oscillation problem
x’’(t) +9x(t)f(t).
~Discuss whether or not resonance occurs.
j~,
,,
4
)‘.
~
(5points)
9
~
2k)
Find e general solution
this forced oscillation problem. Hint: Use the Fourier series forf(t)
given in problem
ou may make use of the particular solutions table at the end of the exam.
(10 points)
,c+
IT
I
1... ~
>cl-9)c~
1tk2
kM (
—
r
~
iç
-1
-L
flkt
cj4~~’
r ~t3
r
)(
≤O;
.fn
•2
7C~
t
1~,
7t
2-
‘C.
it:
çSi’~t
2E
~aa
-i~LoSLt
II
1€
z
-b
$Lkt3
&
It
L
V’3
k ecU~.
v4
3
13
~)
General principles: Pick 3 of the following 4 parts to solve. You will receive credit for the best 3
solutions, for a possible total score of 15 points.
~ Suppose that v is an eigenvector of A, with eigenvalue 2~. Veri~,’ that x( t)
~.
=
St tv is a solution to
6
(5 points)
L
~ Suppose that z is an eigenvector of A, with eigenvalue A. Suppose A < 0 and define w =
Verif~’ thatx(t) = cos(w t)vand4t) = sin(co t)vare solutions to
x”(t)=Ax
(5 points)
-j
x
V
—~
~c~’
V
V
~s
~ coct.oLAj~
14
Prove that if L: V—* W is a linear transformation, and ify~ e V solves the nonhomogeneous equation
L(y~)
f
then every solution of the equation
is of the form y =
L(y) “f
+ y,~ whereyH is some solution of the homogeneous equation
L(y)0.
L Liv)
(5 points)
L (.~, ~
I”
L~
L(~)+ L&h)
4j~’~
LLjY~4.
OSSb ftsrf~l_ It
~fl
~,,
(L~&~—)
~l~0
LCf~~.f
~ Z714L~_yr)
a~,aççL~~
‘
Li. )~L(yp)444—~?) (
~
~
~çj) We discussed the analogy between constant coefficient first-order linear diffe’fential equation4 (ii?
‘ ‘~i Ms
Chapter 1), and first order systems of differential equations (In Chapter 5). Use matrix exponentials and
the “integrating factor” technique to show that for first order systems with constant matrix A, the general ~ a. Ytn
solution to
~‘(t)=A~ +f(i)
is given by the formula
~°
x(t) = e
(A
(Je’~x(t) dt) +
(In the formula above, Je~~’1((t) di is standing for any particular antiderivative ofe”~flt), and the
displayed formula is expressing aa(t) as
+
: begin by rewriting the system as
x’(t)-Axf(t)
and then find an appropriate (matrix) integrating factor.
—)
(5points)
F
2€
~
-
SI
-M
C
A~ =~t
e €
S~
c~
Kit)
(5~ku
~
-
l4-)cft
(f€_A
~
6lt~t
14t
-2
-~~#e
(€44 ~i)
~A ~e~5RJ{
15
Fourier series information: Forf(t) of period P = 2 L,
f—-s
+
21a~cos(n-?t]
+ 2b,,sin(n-~_t)
with
lrL
a0 =
a0
f(i)dt
~=
b
(so
~
-~i--j
—~— =
(~ ~ J)=
(f
~))
sin(n~
f(t)dI is the
average value off)
ijL(~
tJ
dt, naN
rff(t)sin(n~
tJ
dt, tie N
Particular solutions from Chapter 3 or Laplace transform table:
x’’(t) +o~x(t)=Asin(wt)
A
x~(t)
2
2 sin(w t)
when 0) ~
=
0)0
—
A cos(o)ot)
whenoro0
x”(t) +0)~x(t)Ac0s(wt)
A
‘U)
x(1)
2
(00
2 cos(co
r)
when o ≠
A sin(wo
r)
when w=w~
—
=
x”+cx’+w~x=Acos(wt)
x(t)
x(t)
C cos(w t
—
c>O
a)
with
A
cos(a)=
sln(a)
_________
I
2
22 +c2o 2
=
C0)
/12
2\2 +C2 2
x”+cx’+o~x=Asin(wt)
x~(t)
x(t)
C sin(w t
—
c> 0
a)
with
A
//2
2\2 +C20) 2
16
2
2
(O~
(I)
cos(a) =
//2
sin(c*)
2\2
+C20)
2
cw
=
_____________________________
1/2
2\2
+c20
2
17
Table of Laplace Transforms
This table summarizes the general properties of Laplace transforms and the Laplace transforms of particular functions
derived in Chaptcr 10.
f(s)
F(s)
Cs,
of (s) + hg(s)
aF(s) + hO(s)
tfl?’
f’(s)
sF(s)
f”(r)
s2F(s)
—
sf(O)
f(~)(s)
s5F(s)
—
s”1f(0)
f
f(0)
—
(s
cosk
—
[(0)
— .. . —
sinks
f°’”(O)
coshkt
-r~k2
s
f(r)dr
s—k
F(s —a)
t’ coski
(s —a)’+k2
(s
a)f U
—
a)
C°’F(s)
C’ sink!
f(r)g(i
—
r) sir
F(s)GG~
~(sin ks
if (5)
—F’(s)
~sinks
r’f(s)
(—1)’ F’”’(s)
~(sin ki ± k, cot kr)
I
u(s—a)
u(s
f
—
I
f(s).
is
period p
~
F(a)dc
~ fe”f(z)dt
—
ks cos ki)
(~2
-
a ± k2
0)2
-f
—
5
6(5
—
(SI)
a)
a
(square wave)
I
—
s
S
I
s
—
tank
as
—
2
(staircase)
a
s(l—r”)
4
fla ± I)
18