Differential equations
Math 217 — Spring 2010
February Exam
This exam contains fourteen problems numbered 1 through 14. Problems 1 –
13 are multiple choice. Problem 14 is free-response.
Problem 1
A spacecraft is in free fall near the surface of the moon at a speed of 1000
km /hour. Its retrorockets, when fired, provide a constant deceleration
of 20 km/h2 . At what height (in km) above the lunar surface should
the rockets be fired in order to insure a soft touchdown? Treat the
gravitational force of the moon.
A) 1000
D) 10,000
B) 2500
C) 5000
E) 25,000
F) 100,000
Soln: The velocity of the rocket (away from the moon) at time t units
after the retrorockets are fired is 20t − 1000. Integrating gives the distance
d(t) = 10t2 − 1000t + C, where C is the initial distance from the surface. In
order for a soft landing, one wants v(t) = 0 = 20t − 1000 so that t = 50 hours.
If d(50) = 0 as desired, then 0 = 10 · (50)2 − 1000(50) + C or C = 25, 000
km.
1
Problem 2
The fundamental existence and uniqueness theorem states that if both
f (t, y) and ∂f
∂y (t, y) exist and are continuous on a square containing the
point (a, b) in its interior, then for some time interval containing a, the
initial value problem
dy
= f (x, t),
dt
y(a) = b
has one and only one solution defined for t in the interval I.
Which of the following initial value problems is guaranteed to have a
unique solution near the initial value?
I
:
dy
dt
+ ty = t + 1t , y(1) = 0
II
:
dy
dt
+ ty = t + 1t , y(0) = 1
III :
dy
dt
= y 1/3 ,
y(0) = 0
A) Only I B) Only II C) Only III D) I & II E) I & III F) II & III
Soln: For I, f (t, y) = −ty + t + 1t which is continuous whenever t 6= 0. In
addition, ∂f
∂y (t, y) = −t is continuous everywhere. Since the continuity criteria
both apply at (t, y) = (1, 0), there exists a unique solution passing through
(1, 0). For II, f is the same and is discontinuous at t = 0 so the FEU theorem
does not guarantee that a solution exists or is unique. For III, we saw in class
that the multiple solutions exist that satisfy y(0) = 0. So FEU guarantees a
unique solution for I only .
2
Problem 3
Which of the following equations is exact
y
x+y
3
I x + x dx + (y 2 + ln x) dy = 0
II (x + arctan y) dx + 1+y
2 dy = 0
III (ex cos y + cot y) dx + (ex sin y + tan x) dy
IV x dy − y dx
A) I & II B) III & IV C) I & III D) II & IV E) I & IV F) all
∂N
Soln: The criterion for exactness is that ∂M
∂y = ∂x where M (x, y)dx +
∂N
∂M
∂N
2
N (x, y)dy = 0. In I, ∂M
∂y = 1/x = ∂x . In II, ∂y = 1/(1 + y ) = ∂x . In III,
∂M
∂N
∂M
∂N
x
2
x
2
∂y = −e sin y + csc y 6= e sin y + sec x = ∂x . In IV, ∂y = −1 6= 1 = ∂x .
3
Problem 4
Which values of x are stable critical points of the autonomous equation
dx
= (x + 2)(x + 1)(x − 1)(x − 2)?
dt
A) −2 & −1 B) 1 & 2 C) −2 & 1 D) −1 & 2 E) −2 & 2 F)
NONE
Soln: The critical points are x = {−2, −1, 1, 2}. The stable ones are those
dx
x0 such that dx
dt > 0 if x is smaller than but close to x0 and dt < 0 if x is
greater than but close to x0 . From the sign chart, we see that f is changing
Table 1: Sign chart for (x + 2)(x + 1)(x − 1)(x − 2)
x + 2 x + 1 x − 1 x − 2 f (x)
x < −2
−
−
−
−
+
−2 < x < −1
+
−
−
−
−
−1 < x < 1
+
+
−
−
+
1<x<2
+
+
+
−
−
2<x
+
+
+
+
+
from negative to positive when x = {−2, 1}
4
Problem 5
2
Taking for granted that the general solution of t2 ddt2y − 4t dy
dt + 6y = 0 is
2
3
of the form y = At + Bt for appropriate constants A and B, find y(2)
where y is the solution that also satisfies y(1) = 2 and y 0 (1) = −1.
A) y(2) = 0
B) y(2) = −10
D) y(2) = 12
E) y(2) = −5
C) y(2) = π
F) y(2) = −12
2
d y
2
Soln: Since y = At2 + Bt3 , one has dy
dt = 2At + 3Bt and dt2 = 2A + 6Bt.
Plugging in y(1) = 2 gives A + B = 2 while plugging in y 0 (1) = −1 gives
2A+3B = −1. This system of equations is solved by taking (A, B) = (7, −5) .
Then y(2) = −12
5
Problem 6
Identify which of the following provides a general solution of the equation
t2
A) y = te−t + C
dy
= ty + y 2
dt
B) y = t2 + t + C
D) y = t/(C − ln(|t|))
C) y = eln |t| + C
E) y = t ln |t| + C
F) y = Ae−2t
Soln: This can be expressed as the homogeneous equation
dy
= g(y/t); g(u) = u + u2 .
dt
Setting v = y/t so that
equation
dv
dt
= − ty2 + 1t dy
dt or
t
dy
dt
= t dv
dt +v then gives the separable
dv
= g(v) − v = v 2
dt
which integrates to
1
=−
v
Z
dv
=−
v2
Z
dt
= − ln |t| + C
t
or t/y = − ln(|t|) + C or y = t/(C − ln(|t|).
6
Problem 7
dy
The Bernoulli equation x dx
− 5y = 3xy 7/3 can be solved by a suitable
transformation into a linear equation. Which of the following linear
equations represents such a substitution?
A)
dv
dx
+ 2 xv = −1
dv
C) 5 dx
+ 20 xv = −7
dv
E) 12 dx
+ 5 xv = 7
dv
B) 7 dx
+ 15 xv = −3
dv
D) 3 dx
+ 20 xv = −12
F) None of the above
dy
dy
dv
= − 43 y −7/3 dx
= − 43 yv dx
. DiSoln: The substitution v = y −4/3 satisfies dx
dy
− x5 = 3y 4/3 which becomes
viding through the equation by xy gives y1 dx
3 dv
− 4v
dx −
5
x
=
3
v
dv
and, multiplying through by −4v yields 3 dx
+ 20 xv = −12
7
Problem 8
Assume that the volume of water V (t) in a tank at time t obeys Torricelli’s law
p
dV
= −a 2gy
dt
where a is the area of a hole in the bottom of the tank and g = 32ft/sec2
is the acceleration due to gravity.
At time t = 0, a plug at the bottom a cylindrical tank 16 ft high is
removed. After one hour, the water in the tank is 9 ft deep. After how
many hours will the tank be empty?
A) t = 0
B) t = 1
C) t = 2
D) t = 3
E) t = 4
F) Never
√
Soln: The other form of Torricelli’s law is A(y)dy/dt = −k y. Because the
tank is cylindrical, A(y) is a constant that we can absorb into the unknown
√
k to get dy/dt = −k y or
Z
Z
−1/2
y
dy = −k
dt
√
or y = −k/2t + C or y = (C − kt/2)2 . Since y(0) = 16 we have C = 4 and
since y(1) = 9 we have (4 − k/2)2 = 9 or k = 2. Then y = 0 when t = 4 .
8
Problem 9
A turkey is removed from an oven when it has reached an internal temperature of 165◦ F. After 20 minutes, the turkey probe reads 150◦ F. The
temperature in the room is 65◦ F. Assuming Newton’s law of cooling, after how many minutes will the turkey read 120◦ (round to the nearest
minute).
A) t = 60
F) t = 30
B) t = 74
C) t = 45
D) t = 84
E) t = 120
−kt
Soln: Newton’s law says dT
. Since
dt = −k(T − T0 ) or T (t) = T0 + Ae
T (0) = 165 and T0 = 65 we get A = 100. Also, T (20) = 150 = 65 + 100e−20k
or 0.85 = e−20k or k = −(ln 0.85)/20. Then to solve for t, we have 120 =
65 + 100e(ln 0.85)t/20 or 0.55 = e(ln 0.85)t/20 or 20 ln 0.55/ ln 0.85 = t. So t ≈ 74
minutes (t = 73.57)
9
Problem 10
Find the solution to the initial value problem
t2
dy
+ t y = 1,
dt
y(1) = 1
Pick the closest estimate of y(2).
A) 1
B) 0
C) 0.85
D) 1.25
Soln:
E) 0.55
F) 0.75
1
dy y
+ = 2,
y(1) = 1
dt
t
t
So with P (t) = R1/t and Q(t) = t12 one has the linear equation with integrating
R
R
factor ρ(t) = e 1/tdt = eln t = t and ρQ = dtt = ln |t| + C. Then y(t) =
1
1
t (ln |t| + C) so y(1) = C. So C = 1 and y(2) = 2 (ln 2 + 1) ≈ 0.85
10
Problem 11
A population of bacteria increases exponentially, its biomass doubling 3
times per hour. Suppose that an antibiotic is introduced and that, by
varying the concentration one is able to kill the bacteria at a rate of C
units mass per hour. This situation is modeled by the equation
dy
= (3 ln 2) y − C.
dt
Find the smallest number such that any larger concentration will lead to
extermination of the bacterial colony when the initial population is 10
units mass of bacteria.
A)30 ln 2 < C
E) 10 ln 2 < C
B) 20 ln 2 < C
F) none exists
C) 30 < C
D) 40 ln 2 < C
Soln: The model is given by the linear equation dy
dt = (3 ln 2) y − C which
dy
has the form dt + P (t)y = Q(t) where P (t) = −3 ln 2 and Q(t) = −C. So
Z
Z
C
R R
− P
P
3 ln 2t
−3 ln 2t
3 ln 2t
−3 ln 2t
y(t) = e
Qe +D = e
−C e
+D = e
e
+D
3 ln 2
C
C
When t = 0 we get y(0) = 3 ln
2 + D = 10 or D = 10 − 3 ln 2 . Then y(t) =
C
C
3 ln 2t
. According to this formula, the population will die out
3 ln 2 + (10 − 3 ln 2 )e
C
if 10 − 3 ln 2 < 0 or 30 ln 2 < C
11
Problem 12
Identify the differential equation whose slope field appears as follows.
Here, t varies along the horizontal axis and y varies along the vertical
axis.
A)
dy
dt
= y2 − t
B)
2
D) dy
dt = y + t
G)
dy
dt
E)
= t2 − y 2
dy
dt
dy
dt
H)
= y2 + t
C)
dy
dt
= y − t2
= y 2 − t2
F)
dy
dt
= y 2 + t2
I)
dy
dt
= t + y2
dy
dt
= t − y2
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Soln: The slope changes from negative to positive around the curves y = ± t
so dy/dt = y 2 − t . Note that the slope is positive for t < 0.
12
Problem 13
Which direction field belongs to the equation y 0 = y sin t − t sin y?
A) Top left B) Top right C) Bottom left D) Bottom right
4
4
3
3
2
2
1
y
1
0
y
0
−1
−1
−2
−2
−3
−4
−3
−2
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−2
0
2
4
6
8
10
2
4
6
8
10
t
3
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0
2
4
6
8
10
t
Soln: The slope should equal zero whenever y = 0 since 0 sin t − t sin 0 = 0.
Only the top left direction field satisfies this criterion.
13
Name:
Student-ID:
Section:
10 – 11
12 – 1
The following problem is a free-response question. You should justify your
answers.
Problem 14
A cascade of two tanks is set up such that tank 1, initially contains 200
gallons of corn syrup and tank 2 initially contains 100 gallons of water.
Suppose that water flows into tank 1 at a rate of 1 gallon per minute,
that the solution in tank 1 flows from tank 1 into tank 2 at a rate also
of 1 gallon per minute, and that the solution in tank two also exits tank
two at a rate of 1 gallon per minute. Assume that the solutions are
completely mixed instantaneously.
a) Find the amount x(t) of corn syrup in tank 1 at time t > 0
b) Find the amount y(t) of corn syrup in tank 2 at time t > 0
c) What is the maximum amount of corn syrup ever in tank 2?
Soln: We have
dx
x(t)
= −
dt
200
dy
x(t) y(t)
=
−
dt
200
100
The equation for x is given by the exponential decay x(t) = Ae−t/200 =
200e−t/200 . Then the equation for y becomes the linear equation
y
dy
+
= e−t/200
dt 100
R
R
R t/200
−t/200
Let ρ(t) = e dt/100 = et/100 so, with
Q(t)
=
e
one
has
Qρ
=
e
dt =
R
1
t/200
−t/100
t/200
−t/200
200e
+ C and one has y = ρ Qρ = e
200e
+ C = 200e
+
−t/100
−t/200
−t/100
Ce
. Since y(0) = 0, C = −200 so y(t) = 200(e
−e
). Then
dy
−t/100
−t/200
−t/100
−t/200
/100 − e
/200) = 0 when 2e
=e
, that is, when
dt = 200(e
t/200
2=e
or ln 2 = t/200 or t = 200 ln 2. At this time, ymax = y(200 ln 2) =
− ln 2
200(e
− e−2 ln 2 ) or ymax = 50 .