Math 241 – Sample Exam 1 Problems

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Math 241 – Solutions to Sample Exam 1 Problems
Math 241 – Sample Exam 1 Problems
Note. This is not a template, and it is not intended to be a substitute for doing practice problems. It is
likely that topics that do not appear among these sample problems will appear on the exam. However, it is
my hope that these problems will remind you of some of the major topics we have discussed and help you
discover some of the concepts that you need to work on before the exam.
1. Solve each of the following differential equations or initial value problems.
dy
(f) x
= y + xey/x
(a) (y 2 + 1)dx = y sec x dy, y(π/2) = 1
dx
(b) (y 2 + yx)dx − x2 dy = 0
(g) xyy ′ = 3y 2 + x2
2
(c) (y + 4x )dx + 2xdy = 0
dy
(h) xexy
+ yexy = 12x2
(d) (6xy 2 + y 2 )dy + (3x2 + 2y 3 )dx = 0, y(1) = 1
dx
dy
dy
+ 2x2 y − 2x2 + xy − x = 0
(i) x
+ 4y = x4 y 2
(e) x
dx
dx
Solution.
(a) Rewriting and separating variables, we have
(y 2 + 1) dx = y sec x dy
Therefore, we obtain
Z
=⇒
y
dy =
y2 + 1
Z
(y 2 + 1) cos x dx = y dy
cos x dx
=⇒
cos x dx =
=⇒
1
ln(y 2 + 1) = sin x + C1
2
=⇒
ln(y 2 + 1) = 2 sin x + C2
=⇒
y 2 + 1 = Ce2 sin x .
y2
y
dy.
+1
Using our initial condition, we also have
y(π/2) = 1
=⇒
2 = Ce2
=⇒
C = 2e−2 ,
so our final answer is y 2 = 2e−2 e2 sin x − 1, or y 2 = 2e2 sin x−2 − 1.
∂N
(b) For this equation, we have ∂M
∂y = 2y + x and ∂x = −2x, so this is not an exact equation. Therefore, we
must use a different approach to solve it. Note that
(y 2 + yx) dx − x2 dy = 0
=⇒
y 2 + yx = x2
dy
dx
=⇒
y 2
x
+
y
dy
=
,
x
dx
so this differential equation is homogeneous. Therefore, using the usual substitutions v = y/x and
dy
dv
dx = v + x dx , our differential equation becomes
v2 + v = v + x
dv
.
dx
We can now solve the transformed differential equation using separation of variables:
Z
Z
dv
1
1
2
2
v +v =v+x
dv
=⇒ v dx = x dv
=⇒
dx =
dx
x
v2
Therefore, since v = y/x, our final answer is y(x) =
=⇒
ln |x| + C = −v −1
=⇒
v=
−x
.
ln |x| + C
−1
ln |x| + C
2
Math 241 – Solutions to Sample Exam 1 Problems
∂M
∂y
(c) Note that this differential equation is also not exact since
(y + 4x2 ) dx + 2x dy = 0
=⇒
2x
dy
= −(y + 4x2 )
dx
so this differential equation is linear with µ(x) = e
1
dy
+
y = −2x
dx 2x
= 1 and
R
1
2x
dx
√
x, so we have
√ d
y x = −2x3/2
dx
=⇒
√
4
y x = − x5/2 + C,
5
=⇒
so, after dividing through by
= 2. Rewriting, we find that
dy
1
+
y = −2x,
dx 2x
=⇒
= e(1/2) ln x =
√ dy
√
1
x
+ √ y = −2x x
dx 2 x
=⇒
∂N
∂x
√
4
C
x, our final answer is y(x) = − x2 + √ .
5
x
∂N
2
(d) In this case, we have M = 3x2 + 2y 3 and N = 6xy 2 + y 2 , so ∂M
∂y = 6y = ∂x , so this is an exact
equation. Therefore, we have
Z
f (x, y) =
(3x2 + 2y 3 ) dx = x3 + 2xy 3 + P (y),
and
Z
f (x, y) =
(6xy 2 + y 2 ) dy = 2xy 3 +
y3
3 .
so we conclude that f (x, y) = x3 + 2xy 3 +
so using our initial condition we have
y(1) = 1
thus, our final answer is x3 + 2xy 3 +
3x3 + 6xy 3 + y 3 = 10.
The general solution is then given by x3 + 2xy 3 +
13 + 2(1)(13 ) +
=⇒
y3
+ Q(x),
3
13
=C
3
=⇒
C=
y3
3
= C,
10
;
3
10
y3
=
, or, if you want to write it in a somewhat nicer way,
3
3
(e) Rewriting, we have
x
dy
+ 4y = x4 y 2
dx
dy
4
+ y = x3 y 2
dx x
=⇒
y −2
=⇒
dy
4
+ y −1 = x3 ,
dx x
so we see that this is a Bernoulli differential equation with n = 2. Letting v = y 1−2 = y −1 , we see that
dv
−2 dy
dx = −y
dx , and we can now substitute as follows:
y −2
4
dy
+ y −1 = x3
dx x
=⇒
−
dv
4
+ v = x3
dx x
Therefore, our differential equation is now linear in v with µ(x) = e
dv
−4
+
v = −x3
dx
x
=⇒
x−4
dv
− 4x−3 v = −x−1
dx
But recall that v = y −1 , so our final answer is y(x) =
Cx4
(f) Rewriting, we have
x
dy
= y + xey/x
dx
=⇒
dv
−4
+
v = −x3
dx
x
=⇒
R
(−4/x) dx
=⇒
= e−4 ln x = x−4 , so we have
d
x−4 v = −x−1
dx
=⇒
x−4 v = − ln |x| + C
=⇒
v = Cx4 − x4 ln |x|.
1
.
− x4 ln |x|
dy
y
= + ey/x ,
dx
x
dy
dv
so this differential equation is homogeneous. Now, if we let v = xy , then dx
= x dx
+ v, and we can
substitute into our differential equation and use separation of variables as follows:
dy
y
= + ey/x
dx
x
=⇒
x
dv
+ v = v + ev
dx
=⇒
x dv = ev dx
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Math 241 – Solutions to Sample Exam 1 Problems
=⇒
=⇒
−e−v = ln |x| + C1
−v = ln(C − ln |x|)
e
−v
dv =
Z
1
dx
x
=⇒
Z
Therefore, since v = y/x, our final answer is y(x) = −x ln(C − ln |x|).
(g) First, note that
xy
dy
= 3y 2 + x2
dx
=⇒
dy
−3
+
y = xy −1 ,
dx
x
so our differential equation is Bernoulli with n = −1. Letting v = y 1−(−1) = y 2 , we see that
so we can simplify and substitute as follows:
dy
−3
+
y = xy −1
dx
x
=⇒
2y
dy
−6 2
+
y = 2x
dx
x
Therefore, our differential equation is now linear with µ(x) = e
−6
dv
+
v = 2x
dx
x
(−6/x) dx
dv
− 6x−5 v = 2x−5
dx
x−6
=⇒
dy
= 2y dx
,
dv
−6
+
v = 2x
dx
x
=⇒
R
dv
dx
= e−6 ln x = x−6 , so we have
=⇒
d
x−6 v = 2x−5
dx
=⇒
x−6 v =
=⇒
1
v = − x2 + Cx6 .
2
2x−4
+C
−4
1
But recall that v = y 2 , so our final answer is y 2 = − x2 + Cx6 .
2
(h) Rewriting, we have
xexy
dy
+ yexy = 12x2
dx
Therefore, since
have
∂M
∂y
=⇒
xexy dy + yexy dx = 12x2 dx
= xyexy + exy =
∂N
∂x ,
(yexy − 12x2 ) dx + (xexy ) dy = 0.
=⇒
we see that this differential equation is exact. Therefore, we
f (x, y) =
Z
(yexy − 12x2 ) dx = exy − 4x3 + P (y),
f (x, y) =
Z
xexy dy = exy + Q(x),
and
so f (x, y) = exy − 4x3 . Therefore, our final answer is given by exy − 4x3 = C. We could also solve for y
1
and write our final answer as y(x) = ln(4x3 + C).
x
(i) Rewriting, we have
x
dy
+ 2x2 y − 2x2 + xy − x = 0
dx
=⇒
x
dy
+ (2x2 + x)y = 2x2 + x
dx
so we see that our differential equation is linear with µ(x) = e
dy
+ (2x + 1)y = 2x + 1
dx
R
(2x+1)
2
d x2 +x e
y = (2x + 1)ex +x
dx
=⇒
= ex
=⇒
=⇒
Finally, dividing both sides by ex
2
+x
dy
+ (2x + 1)y = 2x + 1,
dx
=⇒
2
+x
. Therefore, we have
Z
2
2
ex +x y = (2x + 1)ex +x dx
ex
2
+x
, we see that our final answer is y(x) = 1 +
y = ex
C
.
ex2 +x
2
+x
+ C.
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Math 241 – Solutions to Sample Exam 1 Problems
2. To the right, you are given a graph of a solution to the
differential equation
10
8
′
y + 2y = mx,
6
where m is some constant. Estimate the value of m.
4
2
1
2
3
4
Solution. Since this differential equation is linear, we could find the general solution in terms of m, use the
graph to estimate two initial conditions, and use our solution to solve for m, but who wants to go through all
of that? It is much easier in this case to use the fact that the point (2, 1.5) appears to lie on the curve, and
that y ′ (2) appears to equal 0. Thus, we conclude that when x = 2, we have y ≈ 1.5 and y ′ ≈ 0. Substituting
this information into our differential equation, we obtain
y ′ + 2y = mx
=⇒
0 + 2(1.5) ≈ 2m
=⇒
m≈
3
,
2
so our final estimate of the value of m is 3/2.
3. For each of the following differential equations, decide for which value(s) of the constant c (if any) the
function y(x) = cx is a solution.
(a) y ′′ − 3y ′ + 3y = x − 1
(b) x2 y ′′ − 3xy ′ + 3y = 0
(c) y ′′ − 3y ′ + 3y = 0
Solution.
(a) Substituting y = cx into the differential equation, we obtain the following expressions for the left and
right hand sides of the DE:
Left hand side
=
Right hand side
=
(cx)′′ − 3(cx)′ + 3cx = 0 − 3c + 3cx = 3c(x − 1),
and
x−1
Therefore, y = cx is a solution precisely when 3c(x − 1) = x − 1 for all x. Since this occurs if and only if
c = 1/3, our final answer is that y = cx is only a solution when c = 1/3.
(b) Substituting y = cx into the differential equation, we obtain the following expressions for the left and
right hand sides of the DE:
Left hand side
=
Right hand side
=
x2 (cx)′′ − 3x(cx)′ + 3cx = 0 − 3xc + 3cx = 0,
and
0
Therefore, the left and right hand sides of the differential equation are equal no matter what the value of
c is. Thus, we conclude that y = cx is a solution for all real numbers c.
(c) Substituting y = cx into the differential equation, we obtain the following expressions for the left and
right hand sides of the DE:
Left hand side
=
Right hand side
=
(cx)′′ − 3(cx)′ + 3cx = 0 − 3c + 3cx = 3c(x − 1),
and
0
Therefore, y = cx can be a solution only when 3c(x − 1) = 0 for all x. Since this can only happen if
c = 0, it follows that y = cx is a solution if and only if c = 0.
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Math 241 – Solutions to Sample Exam 1 Problems
4. Dr. Evil wants to convert his fresh water aquarium into a saltwater aquarium for his mutated brackwater
sea bass. At time t = 0, his aquarium contains 8000 liters of fresh water, and he begins pumping seawater
(which contains 3.5 grams of salt per liter) into his aquarium at a rate of 50 liters per minute. At the same
time, he pumps the well-stirred mixture out of the other side of the aquarium at an equal rate of 50 liters
per minute so that the volume of liquid in the tank remains constant. How long will it take for the water in
the aquarium to reach the desired salinity of 1 gram of salt per liter for his sea bass?
Solution. Let x(t) represent the amount of salt in the aquarium, in grams, after t minutes. Then we have
dx
dt
= (Rate Salt Enters Tank) − (Rate Salt Exits Tank)
50 liters 3.5 grams 50 liters x grams
·
−
·
min
liter
min
8000 liters
x
,
= 175 −
160
=
so after rearranging terms, our modeling differential equation is
dx
1
+
x = 175.
dt
160
Note that this differential equation is linear with µ(t) = e
1
dx
+
x = 175
dt
160
=⇒
=⇒
R
(1/160) dt
= et/160 , so we have
d t/160 e
x = 175et/160
dt
Z
et/160 x = 175et/160 dt
=⇒
et/160 x = 28000et/160 + C
=⇒
x(t) = 28000 + Ce−t/160 .
But we also know that x(0) = 0, so we have
x(0) = 0
=⇒
0 = 28000 + C
=⇒
C = −28000,
which means that x(t) = 28000(1 − e−t/160 ).
To finish, we need to find t so that the concentration of salt in the tank is 1 gram per liter. Since the tank
always contains 8000 liters of water, the concentration of the salt in the tank at any time is given by
x(t)/8000. Therefore, we have
x(t)
=1
8000
=⇒
x(t) = 8000
=⇒
28000(1 − e−t/160 ) = 8000
=⇒
1 − e−t/160 =
=⇒
so our final answer is t = −160 ln(5/7) ≈ 53.8 minutes.
2
7
5
t
,
= ln
−
160
7
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Math 241 – Solutions to Sample Exam 1 Problems
5. Match the differential equations
below to their slope fields given to
the right. Give reasons for your
choice in each case.
(i)
(iii)
y
y
4
4
2
2
(a) y ′ = 1 + y 2
0
x
0
x
(b) y ′ = y
-2
-2
(c) y ′ = x
-4
(d) y ′ = x − y
-4
-4
-2
0
2
4
-4
(ii)
-2
0
2
4
(iv)
y
y
4
4
2
2
0
x
-2
0
x
-2
-4
-4
-4
-2
0
2
4
-4
-2
0
2
4
Solution.
(a) Since y ′ = 1 + y 2 ≥ 1 for all y, it follows that the slope at any point in the slope field for this differential
equation must be at least 1. Therefore, since slope field (iv) is the only one of the four that has positive
slope everywhere, our answer must be (iv).
(b) Since y ′ = y, the slope of any solution curve must be positive when y > 0 and negative when y < 0. The
only slope field that has positive slope whenever y > 0 and negative slope whenever y < 0 is (i), so (i) is
our answer.
(c) Since y ′ = x, the slope of any solution curve must be positive when x > 0 and negative when x < 0. The
only slope field that has positive slope whenever x > 0 and negative slope whenever x < 0 is (iii), so (iii)
is our answer.
(d) Since y ′ = x − y, we see that our slope is zero at any point such that x = y. Therefore, because slope
field (ii) is the only slope field that has zero slope everywhere along the line y = x, we conclude that (ii)
is our answer.
6. The population P of fish in a lake, in thousands, is well-modeled by the differential equation
dP
= 0.2P − 0.02P 2,
dt
where t is measured in years.
(a) Assuming that the fish population starts at 1 thousand, find a formula for the number of fish in the
lake as a function of time.
(b) At what rate is the fish population growing when it reaches half of the carrying capacity of the lake?
(c) Now, assume that h thousand fish are to be harvested from this lake each year.
i. Write down a new modeling differential equation for the fish population in the lake.
ii. What is the largest value that h can have to prevent the situation in which all the fish die out,
regardless of the starting population?
iii. For the value of h you found in part (ii) above, does the modeling ODE from part (i) have an
equilibrium solution? If so, what is it, and is it stable or unstable? Justify your answer.
iv. Draw a bifurcation diagram that describes the dependence of the critical points of the ODE from
part (i) on the parameter h.
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Math 241 – Solutions to Sample Exam 1 Problems
Solution. Note that our differential equation can be rewritten as dP/dt = 0.02P (10 − P ), so this is a
logistic equation with M = 10 and k = 0.02.
(a) Since we are given that P0 = 1, we use the textbook’s formula for the solution of a logistic differential
equation to conclude that
10
P (t) =
,
1 + 9e−0.2t
which is our final answer.
(b) The carrying capacity is M = 10, so we are asking for the value of dP/dt when P = 5. Therefore, using
our differential equation, we have
dP
= 0.02P (10 − P ) = 0.02(5)(10 − 5) = 0.5,
dt
so the population is growing at a rate of 0.5 thousand fish per year, or 500 fish per year.
(c)
i. dP/dt = 0.2P − 0.02P 2 − h
ii. We first find the critical points in terms of h:
dP
=0
dt
=⇒
=⇒
0.02P 2 − 0.2P + h = 0
P =
0.2 ±
p
p
0.2 ± 0.08(0.5 − h)
(0.2)2 − 4(0.02)h
=
2(0.02)
0.04
Now, for all the fish to die out regardless of the initial population, we would need to have no positive
equilibrium solutions, i.e., no positive critical points. We can see from the above calculations that
this happens when h > 0.5, so our answer is h = 0.5.
iii. Yes, if h = 0.5, then, by using our answer to part (c) above, we find that the critical population is
given by P = 0.2±0
0.04 = 5 thousand fish. To analyze stability, we pick two test values, P = 0 and
P = 10, to determine the sign of dP/dt (see below left). Using this table, we then construct a phase
diagram (see below right):
P
dP/dt
0
−
10
−
Therefore, the critical point is semi-stable.
iv. Drawing a bifurcation diagram is equivalent to sketching
the curve dP/dt = 0, which by part (c) is equivalent to the
equation h = −0.02P 2 + 0.2P. By completing the square,
we have
h = −0.02(P 2 − 10P ) =
=
−0.02(P 2 − 10P + 25) + 0.02(25)
−0.02(P − 5)2 + 0.5,
so the relevant curve is a parabola with vertex (h, P ) =
(0.5, 5). The resulting bifurcation diagram is shown to the
right.
P
5
P
10
5
0.5
h
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Math 241 – Solutions to Sample Exam 1 Problems
x1
7. Find the general solution of the following system of equations: 2x1
2x1
+ 3x2
+ 5x2
+ 7x2
+
+
+
3x3
4x3
8x3
=
=
=
13
23 .
29
Solution.

1
 2
2


1
3
13
23  →  0 −1
29
0
1
3 3
5 4
7 8


3 13
1 3
−2 −3  →  0 1
3
2
0 0

3 13
2 3 ,
0 0
so x3 is a free variable; let x3 = t. Therefore, since x2 + 2x3 = 3, we have x2 = 3 − 2x3 = 3 − 2t, and
x1 + 3x2 + 3x3 = 13
=⇒
x1 = 13 − 3(3 − 2t) − 3t = 4 + 3t.
Our final answer is therefore as follows:
Solution Set = {(4 + 3t, 3 − 2t, t) : t ∈ R}
8. Consider the system of equations
3x +
6x +
2y
4y
=
=
a
.
b
(a) Give an example of values for a and b so that the system has no solutions.
(b) For which values of the constants a and b is the system of equations consistent?
Solution.
3 2
6 4
a
b
→
3
0
a
2
0 b − 2a
(a) There are many solutions. For example, for a = 1 and b = 1 the system becomes:
1
3 2
,
0 0 −1
which is inconsistent, and therefore has no solution.
(b) The system is consistent when b − 2a = 0. In this case, the system has infinitely many solutions.