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Math 308 Differential Equations Fall 2002
Homework 2 Solutions
1 & 2. See the last page.
3(a). Newton’s second law of motion says that ma = F , and we know a = dvdt , so we have m dvdt = F. One part of the force is gravity, mg. However, we are using the convention that “up” is positive, and gravity creates an acceleration downward, so the gravitational force is really −mg. In part (a), we are told that there is a frictional force Fd(v) that depends on v. We know that this force should act in the direction opposite to the direction of motion, but we are told that Fd(v) < 0 when v > 0, so that is accounted for in Fd. So the total force is F = −mg + Fd(v). Thus Newton’s second law gives us
m
dv dt
= −mg + Fd(v),
or dv dt
= −g + Fd(v) m
This is a first order differential equation for v, the velocity of the object. If the object is released from height y 0 at t = 0, the velocity at that instant is zero. Therefore the initial condition is v(0) = 0.
3(b). In this case, Fd(v) = 0, and the equation is
dv dt
= −g.
We can integrate this to obtain v = −gt + C,
and the initial condition v(0) = 0 gives us C = 0, so
v(t) = −gt.
As t increases, the velocity decreases linearly with t.
3(c). If Fd(v) = −Cdv, the equation for v is
dv dt
= −g − Cd m
v.
This equation for v is separable, so we can following the usual steps to solve it:
Separate (assuming v 6 = −gm/Cd): dv g + (Cd/m)v
= −dt.
Integrate and solve for v(t). I’ll do this one in detail once more (in the following, exp(x) ≡ ex):
m Cd
ln |g + (Cd/m)v| = −t + k 1 ,
ln |g + (Cd/m)v| = −
Cd m
t + k 2 (where k 2 = (Cd/m)k 1 ),
|g + (Cd/m)v| = exp
Cd m
t + k 2
= k 3 exp
Cd m
t
(where k 3 = ek^2 ),
Note that k 3 must be a positive constant, because it comes from exponentiating k 2. Let’s get rid of the absolute value. We have two cases: g + (Cd/m)v > 0, or g + (Cd/m)v < 0. If g + (Cd/m)v > 0, then |g + (Cd/m)v| = g + (Cd/m)v, and the last equation becomes
g + (Cd/m)v = k 3 exp
Cd m t
If g + (Cd/m)v < 0, then |g + (Cd/m)v| = −(g + (Cd/m)v), so we have
g + (Cd/m)v = −k 3 exp
Cd m
t
The only difference between these two cases is the minus sign. Now, k 3 is a positive constant, so we can combine these two cases into one formula
g + (Cd/m)v = k 4 exp
Cd m
t
where k 4 is an arbitrary nonzero constant. And then we observe that if k 4 = 0, we have g + (Cd/m)v = 0, or v = −mg/Cd, which is the equilibrium solution. Thus we can say that k 4 is an arbitrary constant (including the possibility that k 4 = 0). Finally, we solve for v: v(t) = −
gm Cd
Cd m t
where k is an arbitrary constant. (k is just (Cd/m)k 4 , and since k 4 is arbitrary, so is k.) We want the solution where v(0) = 0, so
v(0) = −
gm Cd
gm Cd
Thus the solution to the initial value problem in this case is
v(t) = −
gm Cd
gm Cd
exp
Cd m
t
gm Cd
1 − exp
Cd m
t
As t increases, exp
− C md t
approaches zero, so the velocity approaches −
gm Cd
. (This is the terminal velocity.)
3(d). We now assume that Fd(v) = −Cdv|v|. Before proceeding, let’s use a little intuition to make life easier. If we drop the object, we expect it to fall; the velocity will be zero initially, and then it will become negative. We expect it to stay negative for t > 0. Therefore, for the solution that we are considering, we have |v| = −v, and Fd(v) = Cdv^2. The differential equation is then
dv dt
= −g + Cdv^2 m
Recall that a =
mg Cd , so we have
v =
mg Cd
1 + K 3 exp
gCd m t
1 − K 3 exp
gCd m t
Now we find the solution for which v(0) = 0:
v(0) =
mg Cd
1 + K 3
1 − K 3
= 0 =⇒ K 3 = − 1.
So the solution to the initial value problem is
v(t) =
mg Cd
1 − exp
gCd m t
1 + exp
gCd m t
As t increases, the exponentials in this formula become large (much larger than 1), and the expression in
parentheses approaches −1. Thus, as t increases, v(t) approaches −
mg Cd.^ Again we see that there is a terminal velocity.
3(e). In (c), we have the equation dvdt = −g − (Cd/m)v, so the term (Cd/m)v must have the same units as dv dt , which is an acceleration, with units m/sec
(^2). v has units of m/sec, and m has units of kg, so Cd must
have units of kg/sec for (Cd/m)v to have units of m/sec^2. In (d), the term (Cd/m)v^2 must have units of acceleration, so in this case, the units of Cd must be kg/m.
- The logistic equation is dP dt = k
P
N
P.
The equilibria are P (t) = 0 and P (t) = N. Now assume P 6 = 0 and P 6 = N. We separate to obtain:
dp ( 1 − (^) NP
P
= k dt.
To integrate the left side, we use the partial fraction expansion: ∫ dp ( 1 − (^) NP
P
N − P
P
dp = − ln |N − P | + ln |P | = ln
P
N − P
(I haven’t included the constant of integration; this will be included in the constant on the right side.) So after integrating we have
ln
∣∣^ P
N − P
∣∣ = kt + C 1.
Now solve for P : ∣ ∣ ∣ ∣
P
N − P
∣ =^ e
kt+C (^1) = C 2 e kt,
P
N − P
= C 3 ekt
Note that C 2 is a positive constant, while C 3 is an arbitrary nonzero constant. As in previous problems, the ±C 2 that arises from eliminating the absolute value is absorbed into C 3. With a little algebra, we find solve for P :
P (t) =
C 3 N ekt 1 + C 3 ekt^
We observe that setting C 3 = 0 gives us P = 0, which is one of the equilibrium solutions, so in fact we can allow C 3 to be an arbitrary constant, including C 3 = 0. However, there is no value of C 3 that will give us the other equilibrium solution P = N , so the general solution is given by (2) or
P (t) = N.
A different (but equivalent) form of the solution is
P =
N
1 + C 4 e−kt^
or P (t) = 0,
where C 4 is an arbitrary constant.
5(a). The amount of drug in the bloodstream at time t (in hours) is a(t) milligrams. There are two processes causing the amount to change:
- The drug clears at a rate that is proportional to the amount present. We are told that the pro- portionality constant is 0.2 per hour, so the rate of change of a(t) from the clearing of the drug is − 0. 2 a(t).
- The drug is being added intravenously. The concentration of the drug in the solution is 1.4 mg/
, and the flow rate is 0. 1/hour, so the rate of change of a is (1.4 mg/)(0.1/hour) = 0.14 mg/hour.
Thus the net rate of change of a(t) is − 0. 2 a + 0.14. This gives us the differential equation
da dt = − 0. 2 a + 0. 14.
We are told that there is initially no drug in the bloodstream, so the initial condition is
a(0) = 0.
The differential equation and the initial condition are the initial value problem.
5(b). The differential equation is separable; in fact, we solved a more general equation like this in class. We find the general solution to the differential equation to be
a(t) = 0.7 + Ce−^0.^2 t.
We use the initial condition to find that C = − 0 .7, so the solution to the initial value problem is
a(t) = 0. 7 − 0. 7 e−^0.^2 t.
5(c).
- After one hour, the amount of drug in the bloodstream is a(1) = 0. 7 − 0. 7 e−^0.^2 = 0.1269 mg.
- After one day, a(24) = 0.6942 mg.
- As t → ∞, e−^0.^2 t^ → 0, so limt→∞ a(t) = 0.7.
1.4/11. Here is a plot of the right-hand side of the differential equation in Exercise 6 versus w:
dw dt
vs. w
0
2
4
–2 –1 1 2 3 4
w
There are two equilibrium solutions, w(t) = −1 and w(t) = 3. For w < −1 or w > 3, w(t) is decreasing, while for − 1 < w < 3, w(t) is increasing. Any solution w(t) for which − 1 < w(0) < 3 will increase monotonically toward 3. If a solution crossed w = 3 (which, as we will see in Section 1.5, is not possible for this equation), it would have to have a slope of zero there. The numerical solution computed in Exercise 6 oscillates around the value w = 3, which our qualitative analysis shows is impossible behavior for a solution to the differential equation.
