Name.
I.D. number
.
Math 2250-4
FINAL EXAM
oL
Ti 0N5
April 3, 2012
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. Laplace Transform tables are
included with this exam. In order to receive full or partial credit on any problem, you must show all
of your work and justify 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. Go&d Luck!
problem
score
possible
1
20
2
20
3
15
4
15
5
10
6
10
7
15
8
15
9
30
total
150
of gravity g = 32
1) Suppose that an object moves vertically, subject only to the acceleration
4-
and a
S
on to be up and write
drag force proportional to the object’s velocity. Choose the positivey directi
drag coefficient, the
the
and
mass
y’ (1) = v(t) for the velocity. For particular values of the object’s
differential equation
=-32—.5v
di’
governs the object’s velocity v(t).
v(t) for all solutions to this differential equation. What
Ia) Construct a phase diagram and determine urn
—
is the term for this limiting velocity?
(5points)
—‘LI
6
v
So
.Lic
s
y
lb) Suppose an object is thrown vertically upwards so that its velocit satisfie
.5v
di’
v(O) 20.
Find a formula for v(t).
dv
e
•5 V
(
f
(lOpoints)
—2.
(—3)
€
.Sv’)
.st
e
-
—/4- 4-
“‘.‘-‘—
#
‘i_I
s
I
v(OO
Ic) Assume the heighty(t) of this thrwn object satisfiesy(O)
ii)
0. Find its the maximum height.
(5points)
v(s)o(c
Y°
Jot
:
cJo
-c
v(4O
te. q:
—
=
I
e’
Ss
-
_f
10
2) Consider the following initial value problem, which could arise from an unforced mass-spring
oscillation problem:
x’’(t) +2x’(t) + lOx(t)rzO
x(O) = 2
x’(O)4.
2a) What sort of damping is exhibited in this problem?
(2 points)
zo
2r1V
2
(r r
r÷)
—
r4-
±3’
2b) Solve this initial problem using the methods of Chapter 5, i.e. by using the characteristic polynomial
mhod.
(13 points)
v’)
4
/
-°h
-t
4
C€
Ik3
.(o) 2c)
#
4
23
4 9
çz
2€
Your
2c) Explain what your solution to 2b) has to do with the phase portrait and curve shown below.
in
shown
equations
differential
explanation should include an explanation of what the system of first order
the pplane output has to do with the second order differential equation in this problem.
(5 points)
)
4
xc
-iUIy
v
)<
t2 t1O(O
2
7
I
c’
)(1
[‘j
1f
Xl.o’l,
-!-l4--.
1
S
Done
“(o’)-’1—
y1oZ
So
v€
1-(-%L
0.
“5
4
o-
r2-i
14.1
3a) Find the general solution to the forced mass-spring problem
x’’(t) +2x’(t) + lOx(t)5+ 101.
Notice that you already found the homogeneous solution in problem 2).
(10 points)
4rL
xrP)
Qi-2c +lv(A4--L) S1-LOE
I,
X
c?
5
(BHo)
4-
(op,’ to
(I
to.
:?)
I
1c2
çC
+
3f
C 1
3b) Use Laplace transform to solve a different forcing problem for this configuration. In this problem the
mass is at equilibrium until time I = 2. At this time it is forced with an impulsive force such thatx(i)
satisfies
x’’(I) +2x’(t) + lOx(tJ3ö(t—2)
x(0) =0
x’(O)O.
(5 points)
s
X()
2cX ()
‘I’
-
AI%sr-
E
—--r
(-I’)
1
ti
(bJ
F(s)
e
r(s)
e k
(5H)
i.e.
e
Ot2.
(1. —2.)
4
4) Consider the forced oscillator initial value problem
0 cos (w
x’’ (1) + o x(t) = F
t)
0
x
x(O)
x’ (0)
=
Assume w * con, i.e. the non-resonance case. Use Laplace transforms to solve the initial value problem.
(15 points)
1’,,’
xo
-
-
ç (s)
i
V
z
F
S
I
X(s
F
-
“
X(c
Fs
0
sx
I
0
/f
s
,
L41
C)
I
4
—
[
I
L4.—CA)
0
[
[
j
-
+
0
X
Li)
0
5
5) Find a basis for
2
consisting of eigenvcctors for the matrix
-4 2
3 -3
Make sure to check your answers, because you’ll be using them in following problems. Hint: the
eigenvalues are integers.
—
-3-3
3
(10 points)
£
#I2—
4-6
—
(2-i)(2i-)
2o
3o
2
2
320
,
4
fci 1R
Lft)L
i
l]1Z1f-2
L iLi L-J
L
2)j
-i
L
1]
L1
/
v
i’
LiJ
6
of differential
6a) Use your work from problem 5) to fmd the general solution to this first order system
equations:
zt(t)
—4 2
ii’ (t)
—
v’(t)
Lvwi
—
3 -3
-tr2
e
1
c
v(t)
(5 points)
4
-•1:
[p3,
matrix, sketch a qualitatively accurate
equations
(5 points)
6b) Using your general solution above and the
picture of the phase diagram for this system of
t
2
l(
7
L
1
t_
4
-
IM’cIcL,\ +t
7et
-S &
5c, L
I,’
.
6
‘S
.fw6’’ d
-3
-2
I’;
+‘(.4
tL
ce’
]
1
vf
Sc7
‘1t’\k._
izt(
-4
,
/
(€.f
2 (1) measure the
1 (t), x
7) Consider the following undamped mass-spring configuration, where as usual x
displacement of the two masses from equilibrium.
t really there), show
1
3 0 (in other words, the third spring isn
2 = 6, k
1 =k
3, m = 2, k
7a) In case
that the resulting system of differential equations reduces to
7
’(t)=-4x +2x
x
1
—
.
2
’(t)3x —3x
1
’
2
x
(
-
3
-
-
xi” -
2x
9x
+
1
‘
“- -
(xi— x
)—
1
(4 poits)
(
x
—
:
2
)
3
L
2
i
X:3,
—
7b) What is the dimension of the solution space to this system of DEs? Explain.
43
s
4s J-4-p 4
Lt lv?
4-
9
c(Jes
(4 points)
i)t
....
oi
‘‘J’-4c
LC L:)
7c) Find the general solution to this sysYem of differential equations. Hint: Use the results of problem 5).
(4 points) Y’(D
2.1F’
3iLxj
L
9
2
r
s4t)) I
Li
;)
)-T
1 1+-)
X
.-
7d) Describe the two fundamental modes of this system.
+t stfrL,
LJ)
(3 points)
5 b+L
L
osc oscRJj
+L
1
O(J
pL
Cç
1’c
2
¶,
7S
I
‘
j
j
8
t for studying second order
8) Although we usually use a mass-spring configuration to give contex
vely exhibit several key ideas from
differential equations, we’ve also used the rigid-rod pendulum to effecti
let the pendulum rod length be L,
,
we
uration
this course. Recall that in the undamped version of this config
pendulum, on which the
assume the rod is massless, and that there is a mass m attached at the end of the
along circular arcs of
vertical gravitational force acts with force magnitude mg. This mass will swing
angle in radians from vertical. The
signed arclength s = L 0 from the vertical reference point where 0 is the
which the mass (and rod)
configuration is indicated below. Note that we are considering a pendulum for
are able to rotate freely about the fixed end of the rod.
-__1
-
the sum of the kinetic energy
8a) The undamped pendulum system satisfies conservation of energy, i.e.
must be constant once the system
from the mass motion plus the potential energy from its change in height
that we’ve studied in this
is put into motion. Use this fact to derive the differential equation for 0(1)
S
(51 “( .l)
course, namely
of m, L, g, 0(t), 0’ (1). Then
Flints: Express the sum of the kinetic energy and potential energy in terms
)
if and only if the time
energy
of
use the fact that the total energy is constant in time (conservation
derivative of the total energy is identically zero.
(8 points)
.-
L (1
6., 4-1--
V
5’(L&(t)
kE
TE
+
vj-
PE
z
—
2-
s—
(7 C?
L
4-
i
v3
-
() L
[L’+s&&]
L
(L’it +
.5
)
/
jL (i— cs)
”+i’&”s
1
L@
n with a linear one when 0(1) was
8b) How did we justify replacing this non-linear differential equatio
0, and what is the linear differential equation we came up with?
near the equilibrium solution 0
(4 points)
1- 2g s
(I,
z. & j. 9—
—
o -(-
E
(I’k4c
‘
- ei
&
I
-
ed period for an oscillating pendulum
8c) Use the linear differential equation in part 8b) to find the expect
in the case of small oscillations, in terms of g, L.
(3points)
f;j
50
.
cI
Y
-
2 yt
9
9) Consider this autonomous system of differential equations for two interacting populations x(t), y(t):
2 +xy
x’(t) x—2x
y’(t) y—y
2 +xy
calling this a “symbiotic logistic populations” model.
in
justified
be
would
someone
9a) Explain why
in these two differential equations, and what they
terms
various
Your explanation should consider the
represent. (The word “symbiotic” means that the presence of each species is beneficial to the other species.
‘c
,—
,
b,
q,
(
i
• +-L
‘
.
S
E’
rr
(“xy’ ( eL
k
c-1 -t-k
(A
points)
ccL)
AJ”f
(i;.
p
9h) Find the four equilibrium solutions algebraically. Plot the cquilibrium points onto the phase portrait
below.
(6 points)
o
yyxyo
7(’r=
°.
0
—
y.y..-:*y
“
7
I I I
1
I
6
(os’)
) (o,o)
1
1
1
I
— —
4
S
—
1
1
1
)(: IL
1
(L4,
o)
4
i
I
.9’ 1-
‘W4
t-2)(4’yzo
-,c= 0
6
1
3
6
7
DorI€
2x—.,,
(V)o)
:
2-?1:3,
—E
1
£
i
(2,
)
I0
System of DEs repeated for convenience:
2
x (t)x—2x +xy
-
xy
y’(t)=y—y
+
2
.
Recall that your choices are: spiral
an matrix to classify each of the e uilbrium
9c Use
our description s ou mclude
source, spiral sink, stable center, nodal source, nodal sink, saddle point
w e er or no e equi i rium point is sta le. Hints: If you do e a gebra correctly and you have the
correct equilibrium points none of the eigenvalues are messy. Also, your answers should be consistent
with the phase portrait on the previous page.
(l6points)
(o,o) , ci I)) (‘fz,o), (2,3)
,c
1
yJ
r-1
10
.::)
2I’)
3
ft -1
-
2
S
i.(
(c3f73)
az
rQ4,o)
-2y
Ly
‘.frLI
L
l_(f2.j
‘2
-
•!
-
--1
=
(
_k)
’1
,(J
1
k
9d) If ye’Jr quiIihrium poi1 nd Icnhian computations above are correct, then you will see that the
linearized system of differential equations at one of the equilibrium points is the same system you studi-....—,—-”
in problem 6). Explain how its solution and the phase portrait you drew in 6) are related to the solutions of
the non-linear system, and to the phase portrait in the non-linear system.
(4 points)
[-
L’i L
yL
iLj
X2 +L.
ç
y)(2,3)
(v)(o,o)
L
p_-f
f’1cL€-L
l’1.x,
:tt,
LL
fr:
t,3)
LAT1L
aAc rfr-
pt