Math 308 Week 2 Solutions
Here are solutions to the even-numbered suggested problems. The answers to the oddnumbered problems are in the back of your textbook, and the solutions are in the Solution
Manual, which you can purchase from the campus bookstore.
Matlab 3
8. Perform the following tasks on the differential equation (1 + t2 )y 0 + 4ty = t.
(a) Use dfield7 to plot a few solutions with different initial points. Start with the
display window bounded by 0 ≤ t ≤ 10 and −5 ≤ y ≤ 5, and modify it to suit
the problem.
(b) Make a conjecture about the limiting behavior of the solutions as t → ∞.
(c) Find the general analytic solution to this equation.
(d) Verify the conjecture you made in part (b), or if you no longer believe it, make
a new conjecture and verify that.
Answer:
(a) The following is a plot of the direction field with the specified bounds (0 ≤ t ≤ 10
and −5 ≤ y ≤ 5) and some solutions:
4
2
y(t)
0
–2
–4
0
2
4
6
t
1
8
10
The following is a plot with display window −5 ≤ t ≤ 5, −5 ≤ y ≤ 5:
4
2
y(t)
0
–2
–4
–4
–2
0
2
4
t
(b) The solutions appear to be limiting to a value slightly above y = 0. (The exact
value, as we shall see in parts (c) and (d) is y = 1/4.)
(c) This is a separable differential equation:
(1 + t2 )y 0 + 4ty
(1 + t2 )y 0
Z
dy
1 − 4y
1
− ln |1 − 4y|
4
| 1 − 4y|
= t
= t(1 − 4y)
Z
t dt
=
1 + t2
1
=
ln(1 + t2 ) + C1
2
2
= e−2 ln(1+t )+C2
1 − 4y = ±eC2 eln(1/(1+t
1
C3
y =
−
4 (1 + t2 )2
2 )2 )
We have to be a little careful in this calculation. When we divided by 1 − 4y
during the initial separation, we lost the solution y = 1/4. (Dividing by 1 + t2 ,
however, did not lose any solutions, since 1 + t2 is always greater than 0). The
lost solution y = 1/4 comes back when we replace ±eC2 with C3 .
Thus, the general solution is
1
C
y(t) = −
4 (1 + t2 )2
2
1
(d) From the solution, we see that as t → ∞, y(t) → .
4
17. A differential equation of the form dx/dt = f (x), whose right-side does not explicitly
depend on the independent variable t, is called an autonomous differential equation.
For example, the logistic model in Example 5 was autonomous. For the autonomous
differential equation
x0 = x(x − 2), −∞ < x < ∞
perform each of the following tasks. Note that the first three tasks are to be performed
without the aid of technology.
(a) Set the right-hand side of the differential equation equal to zero and solve for
the equilibrium points.
(b) Plot the graph of the right-hand side of each autonomous differential equation
versus x. Draw the phase line below the graph and indicate where x is increasing
or decreasing.
(c) Use the information in parts (a) and (b) to draw sample solutions in the xt
plane. Be sure to include the equilibrium solutions.
(d) Check your results with dfield7. Again, be sure to include the equilibrium
solutions.
(e) If x0 is an equilibrium point, i.e., if f (x0 ) = 0, then x(t) = x0 is an equilibrium
solution. It can be shown that if f 0 (x0 ) < 0, then every solution curve that
has an initial value near x0 converges to x0 as t → ∞. In this case x0 is called
a stable equilibrium point. If f 0 (x0 ) > 0 then every solution curve that has an
initial value near x0 diverges away from x0 as t → ∞, and x0 is called an unstable
equilibrium point. If f 0 (x0 ) = 0, no conclusion can be drawn about the behavior
of solution curves. In this case the equilibrium point may fail to be either stable
or unstable. Apply this test to each of the equilibrium points.
Answer:
(a) The equilibrium points are x = 0 and x = 2.
3
(b) Here is the graph of f (x) = x(x − 2) along with the phase line for the differential
equation. We see that f is positive for x < 0 and for x > 2, and that f is negative
for 0 < x < 2. Thus, the solutions x(t) will be increasing for x < 0 and for x > 2,
and that x(t) will be decreasing for 0 < x < 2.
Bœ!
source
Bœ#
sink
If we draw the phase line vertically, as we did in class, we get:
B œ # sink
B œ ! source
4
(c) Based on the phase line, we expect solutions to look something like the following:
Bœ#
Bœ!
The equilibrium solutions are x = 0 and x = 2. Solutions between x = 0 and
x = 2 tend to 0 as t → ∞ and tend to 2 as t → −∞. Solutions with x > 2
increase towards ∞ (we can’t tell from the phase line if these solutions have
vertical asymptotes or not); as t → −∞, these solutions approach 2. Solutions
with x < 0 approach 0 as t → ∞.
(d) Here is a plot of the direction field and some solutions:
4
3
2
x(t)
1
0
–1
–2
–3
–2
–1
0
t
This matches our prediction.
5
1
2
3
(e) In class, we used the terminology sinks and sources instead of stable and unstable. Stable solutions are sinks, and unstable solutions are sources.
We see from the previous parts of this problem that x = 0 is a stable solution (or
sink), and x = 2 is an unstable solution (or source). We can check if it satisfies
the conditions described in (e).
Recall that f (x) = x(x − 2). Thus, f 0 (x) = 2x − 2. Thus, f 0 (0) = −2, and
f 0 (2) = 2. We see that f 0 (0) < 0, and thus x = 0 is a stable equilibrium. We
see that f 0 (2) > 0, and thus x = 2 is an unstable equilibrium.
This gives a different method to determine whether an equilibrium solution is
a sink or source. Instead of determining the sign of f (x) between each of the
equilibrium points, you can compute f 0 (x) at each of the equilibrium points.
However, if f 0 (x) = 0 at an equilibrium point, the equilibrium could still potentially be a sink or a source. Then, you will have to look at the sign of f (x)
below and above the equilibrium point.
This method works for the same reason that the previous method worked. By
finding f 0 (x), you are determining whether f (x) is currently increasing or decreasing at the given equilibrium. If f (x) is increasing, then the sign of f (x)
is changing from negative to positive (remember that f is 0 at the equilibrium
point), and the equilibrium is a source. If f (x) is decreasing then the sign of
f (x) is changing from positive to negative, and f (x) is a sink.
The logistic equation P 0 = rP (1 − P/K) was discussed in class. Usually the parameter’s r
and K are constants, and we found in class that for any solution P (t) which has a positive
initial value we have P (t) → K as t → ∞. For this reason, K is called the carrying
capacity of the system. In the following exercises, you will look at cases where the carrying
capacity is not a constant. In Exercises 23-25, you are to examine the long term behavior
of solutions, especially in comparison to the carrying capacity. In particular:
(a) Use dfield7 to plot several solutions to the equation. (It is up to you to find a
display window that is appropriate to the problem at hand).
(b) Based on the plot done in part (a), describe the long term behavior of the solutions to
the equation. In particular, compare this long term behavior to that of K. It might
be helpful to plot K on the display window. In Exercise 23 the solutions will all be
asymptotic to a constant. In Exercises 24 and 25 the solutions will all have the same
long term behavior. Describe that behavior in comparison to the graph of K.
23. K(t) = 1 − 12 e−t , r = 1. In this case K(t) is monotone increasing, and K(t) is asymptotic to 1. This might model a situation of a human population where, due to technological improvement, the availability of resources is increasing with time, although
ultimately limited.
6
Answer:
(a) Here is a graph of the direction field and some solutions:
5
4
3
P(t)
2
1
0
–1
0
1
2
3
4
5
6
t
There is no constant solution that the solutions are approaching but they do
appear to be approaching y = 1.
(b) Here is a graph of the function K(t):
1
0.8
0.6
0.4
0.2
0
1
2
3
4
5
6
t
As t → ∞, K(t) → 1. The solutions to the differential equation also approach
1 as t → ∞.
7
√
24. K(t) = 1 + t, r = 1. Again K(t) is monotone increasing, but this time it is unbounded. This might model a situation of a human population where, due to technological improvement, the availability of resources is steadily increasing with time,
and therefore the effects of competition are becoming less severe.
Answer:
(a) Here is a graph of the direction field and some solutions:
8
6
4
P(t)
2
0
–2
0
2
4
6
8
10
t
(b) Here is a graph of the function K(t):
3
2.5
2
1.5
1
0
2
4
6
8
10
t
It looks like the solutions to the differential equation approach the function K(t).
Thus, the population tends to infinity as t → ∞.
By the way, in Exercises 23 and 24, the solutions eventually approach K(t) from
below (even the ones that begin above). Whenever y > K(t), the slope of the
8
solution through the point (t, y) will be negative, so the solution will decrease.
Since K(t) is an increasing function (in both 23 and 24) and since the solution
eventually approaches K(t), the solution will eventually need to have a positive
slope. What happens is that the solution passes through the function K(t). At
the point where the solution crosses K(t), it has slope 0 and is a local minima.
After that point, it has positive slope and is increasing towards K(t).
25. K(t) = 1 − 12 cos(2πt), r = 1. This is perhaps the most interesting case. Here the
carrying capacity is periodic in time with period 1, which should be considered to
be one year. This models a population of insects or small animals that are affected
by the seasons. You will notice that the long term behavior as t → ∞ reflects the
behavior of K. The solution does not tend to a constant, but nevertheless all solutions
have the same long term behavior for large values of t. In particular, you should take
notice of the location of the maxima and minima of K and of P and how they are
related. You can use the “zoom in” option to get a better picture of this.
Answer:
(a) Here is a graph of the direction field and some solutions:
5
4
3
P(t)
2
1
0
–1
0
1
2
3
t
9
4
5
6
(b) Here is a graph of K(t):
1.4
1.2
1
0.8
0.6
0
1
2
3
4
5
6
t
If you look closely at these graphs, you will see that the maxima and minima of
K(t) are not the same as the maxima and minima of P (t) (note that the minima
for K(t) occur when t is an integer, and that is not the case for P (t)). It looks
like the maxima and minima for P (t) occur about halfway between the maxima
and minima for K(t) with each maxima for P (t) occurring after a maxima for
K(t), and each minima for P (t) occurring after a minima for P (t).
Here is a graph showing the direction field, one solution, and the graph of K(t).
The graph of K(t) is the curve with the larger amplitude.
3
2.5
2
1.5
P(t)
1
0.5
0
–0.5
–0.5
0
0.5
1
1.5
t
10
2
2.5
3
If you think about what this means for the population where K(t) is periodic due
to seasons, this should make sense. The population does not begin decreasing
immediately after the temperatures begin decreasing in the fall. Instead, the
population continues increasing, just not as quickly as it was increasing before.
A short while later, the population starts decreasing.
NSS 1.4
12. In Example 2 we approximated the transcendental number e by using Euler’s method
to solve the initial value problem
y 0 = y,
y(0) = 1
Show that the Euler approximation yn obtained by using the step size 1/n is given
by the formula
µ
¶n
1
,
n = 1, 2 . . .
yn = 1 +
n
Recall from calculus that
µ
¶n
1
lim 1 +
=e
n→∞
n
and hence Euler’s method converges (theoretically) to the correct value.
Answer: There’s a possible confusion in this problem with n meaning two different
things — the nth step in Euler’s method and the step size 1/n. To simplify things,
we will use k to denote the current step in Euler’s method. We want to show that
the formula for the y value after n steps of step-size 1/n is
µ
¶n
1
yn = 1 +
,
n = 1, 2 . . .
n
Using step size 1/n, the x values between 0 and 1 are 0, 1/n, 2/n, 3/n, . . . , (n−1)/n, 1.
To get the next y-value at each step in Euler’s method, we use the formula
yk+1 = yk + hf (xk , yk )
In the formula, h = 1/n (the step size) and f (xk , yk ) = yk (since we are considering
the differential equation y 0 = y). Using this information, the formula becomes
yk+1 = yk +
This is equivalent to
yk+1
yk
n
µ
¶
1
= yk 1 +
n
This formula looks promising. Using it, Euler’s method gives us the following table:
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µ
¶
1
= yk 1 +
n
k
xk
yk
yk+1
0
0
1
y1 = 1 +
1/n
1
1+
n
2/n
µ
¶2
1
1+
n
3
3/n
µ
¶3
1
1+
n
..
.
..
.
1
2
µ
y2 =
µ
y3 =
µ
y4 =
..
.
µ
¶n−1
1
n − 1 (n − 1)/n
1+
yn =
n
1
n
1
1+
n
1
1+
n
1
1+
n
..
.
µ
1
1+
n
¶2
¶3
¶4
¶n
Thus, we get the formula we were trying to show:
¶n
µ
1
yn = 1 +
n
And, from calculus, we know that
¶n
µ
1
lim 1 +
=e
n→∞
n
Thus, Euler’s method converges to the correct value. (By the way, the above limit is
a relatively important limit, and is often used as the definition of e. There’s not a
standard place for the limit in the Calculus curriculum, though, so you may not have
seen it before.)
NSS 2.2
30. As stated in this section, the separation of equation (2) on page 40 requires division
by p(y), and this may disguise the fact that the roots of the equation p(y) are actually
constant solutions to the differential equation.
(a) To explore this further, separate the equation
dy
= (x − 3)(y + 1)2/3
dx
to derive the solution
y = −1 + (x2 /6 − x + C)3
12
(b) Show that y ≡ −1 satisfies the original equation dy/dx = (x − 3)(y + 1)2/3
(c) Show that there is no choice of the constant C that will make the solution in
part (a) yield the solution y ≡ −1. Thus, we lost the solution y ≡ −1 when we
divided by (y + 1)2/3 .
Answer:
(a) We separate:
Integrating, we get
Z
dy
=
(y + 1)2/3
Z
(x − 3) dx
1
3(y + 1)1/3 = x2 − 3x + C
2
Thus:
y = −1 + (x2 /6 − x + C)3
(b) The constant solution y = −1 is a solution to the differential equation, because
the derivative of the constant function is 0, and if we plug y = −1 into the
equation (x − 3)(y + 1)2/3 we get (x − 3)(−1 + 1)2/3 = 0.
(c) For every value of C, the solution y = −1 + (x2 /6 − x + C)3 is a cubic equation
and not a line. Thus, we lost the constant solution y = −1 when we divided by
(y + 1)2/3 .
When you solve a separable differential equation, you should be careful about
this. Check during the separation step to see if you are dividing by a constant
solution. If you are, then check your final answer to see if it incorporates the
constant solution (sometimes it does). If it does not, then you should make sure
to include the constant solution as part of your final answer.
38. Free Fall. In Section 2.1, we discussed a model for an object falling toward Earth.
Assuming that only air resistance and gravity are acting on the object, we found that
the velocity v must satisfy the equation
dv
m = mg − bv
dt
where m is the mass, g is the acceleration due to gravity, and b > 0 is a constant (see
Figure 2.1). If m = 100 kg, g = 9.8 m/sec2 , b = 5 kg/sec, and v(0) = 10m/sec, solve
for v(t). What is the limiting (i.e., terminal) velocity for the object?
Answer: We have the differential equation
dv
100 = 980 − 5v
dt
This equation is separable:
Z
Z
100 dv
=
dt
980 − 5v
−20 ln | 980 − 5v| = t + C
| 980 − 5v| = e−(t/20)+C
980 − 5v = Ae−t/20
v = 196 − Ae−t/20
13
We have to be a little careful in this calculation. When we divided by 980 − 5v during
the initial separation, we lost the solution v = 196. The lost solution v = 196 comes
back when we replace ±eC with A.
The initial condition v(0) = 10, tells us that 10 = 196 − Ae0 . Solving for A, we get
A = 186. Thus, the solution is
v(t) = 196 − 186e−t/20
From this solution, we can see that as t → ∞, v(t) → 196. Thus, the terminal
velocity is 196 m/sec
14