C Roettger, Fall 13
Math 267 - Practice exam 2 - solutions
Problem 1 A solution of 10% perchlorate in water flows at a rate of 8
L/min into a tank holding 200L pure water. The solution is kept well stirred
and flows out of the tank at a rate of 6 L/min.
a) Determine the volume of perchlorate in the tank after t minutes.
b) When will the percentage of perchlorate in the tank reach 6%?
Solution. a) Let V (t) and x(t) the volume of water and perchlorate in the
tank at time t, respectively. So
V (t) = 200 + 2t
(time in minutes, volume in liters). Then the amount x(t) of perchlorate
satisfies
6x(t)
.
x0 (t) = 0.10 × 8 −
V (t)
This is a linear equation, written in standard form as
x0 (t) +
3
x = 0.8,
100 + t
so an integrating factor is
µ(t) = e
R
3/(100+t) dt
= e3 ln(100+t) = (100 + t)3 .
We multiply the entire equation by µ and integrate:
Z
µ(t)x(t) = 0.8 (100 + t)3 dt = 0.2(100 + t)4 + C.
so after solving for x, we get the general solution
x(t) = 0.2(100 + t) +
C
.
(100 + t)3
Since x(0) = 0, we can solve for C to get
C = −0.2 · 108 = −20, 000, 000.
1
The amount of perchlorate after t minutes is
x(t) = 0.2(100 + t) −
20, 000, 000
.
(100 + t)3
b) We put
0.06 =
x(t)
10, 000, 000
= 0.1 −
V (t)
(100 + t)4
and solve for t, getting first
(100 + t)4 =
107
0.04
and then
t=
p
4
107 /0.04 − 100 ≈ 25.74
(in minutes).
Problem 2 A garage with no heating or cooling has a time constant of 2
hours. The outside temperature varies as a sine wave,
M (t) = 65 + 15 sin(π(t − 8)/12).
Determine the time when the building reaches its lowest and its highest
temperature, assuming that the exponential term has died off.
2
Note how the building temperature T (blue) lags behind the ambient temperature M !
Solution.
equation
The temperature T (t) of the building satisfies the differential
T 0 (t) = k(M (t) − T (t)).
Rewritten in standard form for a linear equation, this reads
T 0 + kT = kM
so we have the integrating factor µ(t) = ekt , and the usual recipe gives
Z
kt
e T (t) = kekt (65 + 15 sin(π(t − 8)/12)) dt.
Integrating by parts twice gives for the hard part on the right-hand side
Z
ekt [k sin(αt + β) − α cos(αt + β)]
+ C.
ekt sin(αt + β) dt =
α2 + k 2
Integrating 65kekt gives another 65ekt . We substitute k = 1/2 (1 / time
constant), α = π/12, β = −2π/3. Altogether,
π(t−8)
π(t−8)
15
5π
sin
− 8 cos
4
12
12
T (t) = 65 +
+ Ce−t/2 .
2
π /144 + 1/4
For t large enough so that the exponential term ’has died off’ (meaning we can
neglect it), we can find the maximum / minimum temperature by setting the
derivative of the first term equal to 0, like in Calculus I. It helps to introduce
a new variable u = π(t − 8)/12. So we differentiate
d 15
5π
15
5π
sin u −
cos u =
cos u +
sin u = 0
du 4
8
4
8
which gives
6
+ nπ ≈ −1.088 + nπ
u = − arctan
π
for any integer n. Going back to t,
π(t − 8)
= −1.088 + nπ
12
3
gives the times
t = 12.158 + 12n
for any integer n. A second-derivative test (no, we don’t want that) or firstderivative test (still, lots of work) or simply a graph (best! see above) shows
that even integers n give times with minimal temperature, odd integers n
give maximal temperature.
Problem 3 A parachutist whose mass is 75 kg drops from a helicopter
hovering 2000 meters above the ground. She falls to the ground under the
force of gravity and air resistance, which is proportional to her velocity, with
proportionality constant b1 = 30 N sec/m when the chute is closed, and
b2 = 90 N sec/m when the chute is open. The chute does not open until
the velocity reaches 20 m/sec. At what time will she reach the ground? Use
g = 10N/kg.
Solution. Let y(t) be the parachutist’s height above ground. So y(0) =
2000, y 0 (0) = 0, and
my 00 + by 0 = −mg
We write v = y 0 , so we get a first-order equation for the velocity,
v0 +
bv
= −g
m
which has integrating factor µ = ebt/m ,
Z
gmebt/m
vµ = −g ebt/m dt = −
+C
b
and the solution
gm
+ Ce−bt/m .
b
This analysis is valid both for open and closed parachute. Now use v(0) = 0
to get for the first, rapid fall
v=−
C1 =
gm
= 25,
b1
v = 25(e−b1 t/m − 1).
Let us find the time t1 when the parachute opens, so v = −20 (negative sign
because the movement is down!!)
−20 =
gm −b1 t/m
(e
− 1)
b1
4
gives with g = 10N/kg
20b1
m
≈ 4.02
t1 = − ln 1 −
b1
gm
(in seconds). After that time, we get
v=−
gm
+ C2 e−b2 t/m .
b2
and the initial condition v(t1 ) = −20 gives
C2 = −1458.33
Finally, find the position y(t) by integrating v. In both cases,
y(t) = −
gmt Cm −bt/m
−
e
+ D.
b
b
For the first case, y(0) = 2000 gives
D1 = 2000 +
C1 m
= 2062.5
b1
so the position at time t1 is y(t1 ) ≈ 1949.41 meters. With the parachute
open, we get
gmt C2 m −b2 t/m
y(t) = −
−
e
+ D2
b2
b2
and from the initial condition y(t1 ) = 1949.41, we can determine
D2 = 1973.22
Finally, the parachutist reaches the ground when
0 = y(t) ≈ −
so at time
t=
gmt
+ D2
b2
D2 b2
≈ 236.8
gm
(in seconds). The exponential term at around this time is completely negligible.
5
Note. This is 3.4 # 7, and the book’s answer is 241 seconds, because they
used the more accurate value g = 9.81N/kg.
Problem 4 Find the general solution of
y 00 − 2y 0 + 5y = 0.
Solution. The auxiliary equation is
r2 − 2r + 5 = 0
which has roots 1 ± 2i. So we can write the general solution as
y(t) = et (c1 cos 2t + c2 sin 2t).
Problem 5 Consider the differential equation
x00 (t) − 2x0 (t) + x(t) = cos t.
(1)
a) Find a particular solution of equation (1).
b) Find the complimentary function (general solution of the associated homogeneous equation).
c) Find the unique solution satisfying both (1) and the initial conditions
x(0) = 0, x0 (0) = 1.
Solution. a) We try xp (t) = A cos t + B sin t. This leads to A = 0, B =
−1/2.
b) The characteristic equation is (r − 1)2 = 0, with double root r = 1, so the
complimentary function is
xc (t) = (c1 + c2 t)et
c) We get c1 = 0, c2 = 3/2 after substituting the general solution xp + xc for
x.
Problem 6 A mass of 2 kg is traveling horizontally on wheels without
friction. It is hooked up to a spring with spring constant k = 162 N/m. Let
x(t) be the position of the mass with x = 0 being the equilibrium position.
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a) Find the second-order linear differential equation governing x.
b) What is the natural frequency ω of this system?
c) Find the position function x(t) if x(0) = 12 and x0 (0) = −45.
d) Suppose an external force F = 2 cos(10t) starts acting on the mass, and
x(0) = x0 (0) = 0. Find the resulting position function x(t).
Solution. a)
2x00 + 162x = 0.
p
b) ω = 162/2 = 9 (in radians per second).
c) The general solution here is
xc (t) = c1 cos 9t + c2 sin 9t.
The initial conditions give c1 = 12 and c2 = −5, so
x(t) = 12 cos 9t − 5 sin 9t.
d) Now we have to find a particular solution xp for
2x00 + 162x = 2 cos 10t.
(2)
There is no overlap between cos 10t, sin 10t and cos 9t, sin 9t., So we try
xp (t) = A cos 10t + B sin 10t,
and substituting this into (2) yields equations B = 0, −200A + 162A = 2, so
A = −1/19. The general solution is now
x(t) = −
1
cos 10t + c1 cos 9t + c2 sin 9t.
19
The initial conditions give x0 (0) = 9c2 = 0, so c2 = 0 and c1 = 1/19.
Problem 7 Consider the Euler-Cauchy equation
t2 y 00 + 3ty 0 + y = 0.
a) Find the general solution.
b) Then find the solution satisfying y(1) = 4, y 0 (1) = 3.
Solution. The characteristic equation here is
r(r − 1) + 3r + 1 = 0
7
which has a double root r = −1, so solutions are y = 1/t and y = ln(t)/t.
The general solution is
c1 + c2 ln t
.
y=
t
b) So c1 = 4 and from
y 0 (t) =
c2 /t − (c1 + c2 ln t)
,
t2
we get c2 − c1 = 3, so c2 = 7.
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