MA 226, Summer I 2014 Midterm 1

MA 226, Summer I 2014
Midterm 1 - Solutions
Name:
For full and partial credit please show all work below. You do NOT need a calculator for any of
the following problems, and all answers can be left unsimplified if necessary.
PROBLEM 1 (10 points). Use the following graph of f (y) to sketch the phase line of dy/dt =
f (y). Also, identify any sinks, sources, and/or nodes. Finally, sketch the general solutions of the
differential equaiton dy/dt = f (y).
y
y
f (y)
3
2 sink
1
−3 −2 −1 0
1
2
3
y
0 node
−3 −2 −1 0
1
2
3
t
−2 source
PROBLEM 2 (10 points). Find the general solution to the differential equation dy
= −2ty 2 . Then,
dt
solve the initial value problem with y(0) = 2.
Solution. From first glance, we see that this differential equation is separable. Hence, we can solve
it by
dy
dt
1 dy
y 2 dt
1
dy
y2
−1
y
= −2ty 2
= −2t
−2t dt
=
= −t2 + c
y=
−1
+c
t2
for an arbitrary constant c. Taking all values of c gives us the general solution to the given differential equation, except we’re missing one solution. Just by inspection, we have y = 0 is a solution
as well, which is not one of the solutions given by our method of separation. Now, to solve the
initial value problem, we first note that y = 0, so we have
y(0) = 2 =
−1
1
=− .
2
−0 + c
c
1
MA 226, Summer I 2014
Midterm 1 - Solutions
Name:
Therefore, c = − 12 , and the particular solution to this initial value problem is
y(t) =
t2
1
+
1
2
=
2t2
2
.
+1
PROBLEM 3 (20 points). Consider the following 8 first-order equations:
dy
dy
dy
dy
=t−1
2.
=t+1
3.
=y+1
4.
=1−y
dt
dt
dt
dt
dy
dy
dy
dy
5.
= y2 + y
6.
= y(y 2 − 1)
7.
=y−t
8.
=y+t
dt
dt
dt
dt
Four of the associated slope fields are shown below. Pair the slope fields with their associated equations. Provide a brief justification for your choice.
1.
Slope Field A
3
Slope Field B
3
2
2
1
1
-3 -2 -1
-1
1
2
-3 -2 -1
-1
3
-2
-2
-3
-3
Slope Field C
3
2
1
1
1
2
2
3
Slope Field D
3
2
-3 -2 -1
-1
1
-3 -2 -1
-1
3
-2
-2
-3
-3
2
1
2
3
MA 226, Summer I 2014
Midterm 1 - Solutions
Name:
Solution. The solution for each slope field is as follows:
A) 7 - Our first observation might be that the differential equation here must involve both y and t,
since the slopes are different along horizontal lines as well as vertical lines. Next, it appears
that the slope of the slope marks is 0 along the line y = t. This suggests that the differential
equation in question should have a y − t term. Indeed, the one such equation is given by 7.
B) 2 - Because the slope marks are parallel along vertical lines, we know that the differential
equation must be dependent only on t. Furthermore, the slope is 0 when t = −1. The only
equation with these properties is 2.
C) 3 - From the parallel slopes along horizontal lines, we see that the differential equation we want
is autonomous. Also, there is just one equilibrium solution, which occurs at y = −1. While 5
and 6 are autonomous with equilibrium solutions at y = −1, they have additional equilibrium
solutions. Therefore, the correct differential equation is 3.
D) 5 - Similar to the last slope field, the differential equation here is autonomous and has equilibrium solutions at y = −1 and y = 0. Since we’ve already ruled out 3 and 6 has an equilibrium
solution at y = 1, we know that the differential equation has to be 5.
PROBLEM 4 (20 points). In the following parts, you will find the general solution to the differential
equation
−ty
1 + 2t2
dy
=
+
dt
1 + t2
1 + t2
a) (10 points) Solve the associated homogeneous equation
dy
dt
=
−ty
.
1+t2
Solution. Recall that to find the general solution of a nonhomogeneous equation, we just need
one solution, yh , to the associated homogenous equation (by the extended linearity principle).
Because every homogeneous first order linear differential equation is separable, we can solve
by separation. We have
−t
1 dyh
=
yh dt
1 + t2
1
−t
dyh =
dt
yh
1 + t2
1
ln yh = − ln(1 + t2 )
2
1
yh = √
1 + t2
which is well defined for all t.
3
MA 226, Summer I 2014
Midterm 1 - Solutions
Name:
b) (5 points) Find a particular solution to the given nonhomogeneous differential equation. [Hint:
Stare at the equation hard enough, and try to guess a particular solution.]
Solution. We just need one particular solution to our given differential equation, so we can
simply guess and check that our guess is indeed a solution. A natural first guess is y equal to a
constant. However, this clearly will not work. The next guess you might try is y = αt. In fact,
if we just try y = t, we get
1=
−t2
dy
1 + 2t2
1 + t2
=
+
=
= 1,
dt
1 + t2
1 + t2
1 + t2
which is true. Therefore, yp (t) = t is a particular solution to our problem.
c) (5 points) What is the general solution? Justify your answer.
Solution. By the extended linearity principle, the general solution to a nonhomogeneous equation is given by y(t) = kyh (t) + yp (t), where k is an arbitrary constant. Therefore, using parts
a) and b) we find the general solution is given by
y(t) = k(1 + t2 )−1/2 + t.
PROBLEM 5 (10 points). Using the method of “guessing,” solve the initial value problem
dy
+ 5y = 3e−5t ,
dt
y(0) = −2.
Solution. Using the method of guessing we discussed in class, our first guess of a particular solution
would be yp (t) = αe−5t . However, this guess doesn’t work, since αe−5t is already a solution to
the associated homogeneous equation dy/dt + 5y = 0. Therefore, we make a second guess of
yp (t) = αte−5t . We have
dyp
+ 5yp = −5αte−5t + αe−5t = 5αte−5t = 3e−5t ,
dt
which is true if α = 3. Therefore, one particular solution is yp (t) = 3te−5t . Now, using the extended linearity principle, we obtain the general solution, y(t) = ke−5t + 3te−5t for some constant
k. Now, using the initial value y(0) = −2, we solve for k and find k = −2. Therefore, the solution
to our initial value problem is y(t) = −2e−5t + 3te−5t .
PROBLEM 6 (10 points). Justify your answer in Problem 5 using an integrating factor.
Solution. The first step is to find the integration factor µ, since the differential equation is already
written in the form dy/dt + g(t)y = b(t). Using the formula we developed in class, we have
µ(t) = e
g(t) dt
=e
4
5 dt
= e5t .
MA 226, Summer I 2014
Midterm 1 - Solutions
Name:
Now, we multiply both sides of the differential equation and use the product rule to get
e5t
dy
+ e5t 5y = e5t 3e−5t
dt
d(e5t y)
=3
dt
d(e5t y)
dt = 3 dt
dt
e5t y = 3t + c
y = 3te−5t + ce−5t
which is the same as the general solution from Problem 5. Hence, we can solve the rest of the
initial value problem the same way to get the solution y(t) = −2e−5t + 3te−5t .
PROBLEM 7 (20 points). A cup of coffee is initially 150◦ . Suppose it cools to 140◦ in 1 minute in
a room with an ambient temperature of 70◦ .
a) (10 points) Assume that Newton’s law of cooling applies: The rate of cooling is proportional to
the difference between the current temperature and the ambient temperature. Write an initialvalue problem that models the temperature of the coffee.
Solution. Let y(t) be the temperature of the coffee cup in degrees Fahrenheit at time t, where t
is in minutes. Also, let k < 0 be the rate proportionality between dy/dt and y − 70. Then, the
temperature of the coffee cup can be modeled by the differential equation
dy
= k(y − 70).
dt
Note that we need k < 0 in order for our model to be decreasing or “cooling” for y > 70.
To make an initial value problem out of our model, we need to specify an initial condition,
y(0) = 150. However, if we were to carry out this problem, we need to also know k, in which
case we also need y(1) = 140.
b) (10 points) What happens to the temperature of the coffee as time approaches infinity? Justify
your answer either by using a phase line or solving the initial value problem you wrote in part
a) and taking the limit as t → ∞.
Solution. If we find the equilibrium solutions to our differential equation, we notice there is
d
only one equilibrium solution at y = 70. Also, we have dy
(k(y − 70)) = k < 0 for all y.
Therefore, y = 70 is a sink by linearization. Hence, no matter what the initial temperature of
the coffee cup is, it will always asymptotically approach the ambient temperature, 70◦ .
5

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