Download Solutions to Math 308 Homework 1: Differential Equations and Radioactive Decay and more Assignments Differential Equations in PDF only on Docsity!
Math 308 Differential Equations Fall 2002
Homework 1 Solutions
Text Problems
(a) The statement of the problem says t is time, but it does not specify the starting point. I will define t to be the number of years since 1939. Then the initial condition, given by the first entry in the table, is A(0) = 32800. As shown in Section 1.1 (and subsequently shown in class), the solution to the differential equation is
A(t) = A 0 ekt,
where A 0 is a constant. To find A 0 , we use the initial value:
A(0) = A 0 e^0 = A 0 = 32800.
so the solution to the initial value problem is
A(t) = 32800ekt.
(b) If the data agrees with the model, we can use any data point (except for the first) to determine k. I will try the model with two values of k, determined by two different data points. To find k 1 , I will use the second data point: A(5) = 55800. Thus
A(5) = 32800e^5 k^1 = 55800 =⇒ e^5 k^1 =
=⇒ k 1 =
ln
To find k 2 , I use the last data point, A(35) = 584000. We find
k 2 =
ln
(c) Here is a table that shows that actual data, and the predictions of the model with k = k 1 and k = k 2. Only three significant digits are shown for the predictions. Predicted Predicted Year t Actual with k = k 1 with k = k 2 1939 0 32,800 32,800 32, 1944 5 55,800 55,800 49, 1949 10 73,600 94,900 74, 1954 15 138,000 161,000 113, 1959 20 202,000 275,000 170, 1964 25 257,000 467,000 257, 1969 30 301,000 795,000 387, 1974 35 584,000 1,350,000 584, 2010 71 6. 20 × 107 1. 13 × 107 2050 111 4. 35 × 109 3. 03 × 108 2100 161 8. 84 × 1011 1. 86 × 1010
(d) The following graph compares the actual data with the models for 0 ≤ t ≤ 35.
0 5 10 15 20 25 30 35 0
2
4
6
8
10
12
14 x 10^5
Actual Predicted, k=k 1 Predicted, k=k 2
With k = k 1 , the predictions are significantly larger than the the actual data. The model does not appear very good, and it would not be prudent to believe the predictions for 2010, 2050, and 2100, especially since the model gives areas that exceed the area of Australia! With k = k 2 , the model appears more reasonable for the time 0 ≤ t ≤ 35. However, this model also predicts that the area will exceed the area of Australia by 2010, so we still can’t believe the predictions. It appears that the equation dAdt = kA is not a good model. A model based on the logistic equation might be better.
(a) We will use the notation given and translate the verbal description of the behavior into a mathematical equation. First, “the rate at which a quantity of a radioactive isotope decays” refers to drdt ; the word “decay” suggests that we expect the rate of change of r(t) to be negative. The description says that the rate of change is proportional to the amount present, which is just r(t). The proportionality constant is what the problem statement calls “−λ”. Thus the equation is
dr dt = −λr(t).
(b) The statement “the amount of the isotope present at t = 0 is r 0 ” translates directly into r(0) = r 0 , so the initial value problem is dr dt
= −λr(t), r(0) = r 0.
1.1/9. First, some general comments about half-life. We know that the amount of an isotope is determined by the equation dr/dt = −λr(t). The solution to this first order differential equation is r(t) = r 0 e−λt, where (as in the previous problem) r 0 is the amount present at t = 0. Let T be the half-life of the isotope;
Let’s plot dP/dt vs. P :
dP/dt vs. P
0
50 500 1000 1500 P 2000 2500 3000
The equilibria are P ≈ 396 and P ≈ 2104. For P > 2104, we see that dP/dt < 0, which tells us that P will decrease. In fact, P (t) will continue to decrease towards the equilibrium at P ≈ 2104. In other words, in the long-term we expect the population to level off at about 2104.
(b) In this case, the model is
dP dt
= k
P
N
P −
P
P
P −
P
and the plot of dP/dt vs. P is:
dP/dt vs. P
0
200 500 1000 P 1500 2000 2500
The equilibria are P = 0 and P = − 2500 / 9 ≈ −278. Of course, negative values of P are not relevant for the model, but this does tell us that dP/dt < 0 for all P > 0. This means that for any starting population P (0) (including the case P (0) = 2500), P (t) will decrease towards zero. In the long-term, the population will die out.
Separate: y dy = t dt Integrate: y^2 2
t^2 2
+ C
Solve for y(t):
y = ±
t^2 + 2C.
Note that this means there are two families of solutions: y =
t^2 + 2C and y = −
t^2 + 2C. If an initial condition were given, it would determine the sign of the solution, and the value of C.
We have dy dt
ty + t + y + 1
(t + 1)(y + 1)
t + 1
y + 1
Now separate:
(y + 1) dy =
dt t + 1
Integrate: y^2 2
Solve for y:
y^2 2
y = − 1 ±
1 + 2 ln |t + 1| + 2C 1
or y = − 1 ±
2 ln |t + 1| + C 2
where C 2 = 1 + 2C 1.
We have dy dt = ty^2 + 2y^2 = (t + 2)y^2
Separate: dy y^2
= (t + 2) dt
Integrate:
−
y
t^2 2
We can use the initial condition y(0) = 1 (that is, y = 1 when t = 0) to determine C:
= 0 + 0 + C =⇒ C = − 1
So we have
−
y
t^2 2
Solve for y:
y =
t^2 2 + 2t^ −^1
t^2 + 4t − 2
(c) Solving the equation for v is certainly easy enough:
v(t) = 10t + C.
The differential equation for r is separable, and we could following the usual procedure to solve it. Or, we can solve v = 4πr^3 /3 for r:
r =
3 v 4 π
Now put in v(t):
r(t) =
3(10t + C) 4 π
This is the solution to the differential equation for r.
(a) The equation for a(t) is dadt = −ka, and by now we know that the solution is a(t) = a 0 e−kt, where a(0) = a 0. We are told that the patient is given 5 units at t = 0, so a(0) = 5. Therefore, a(t) = 5e−kt. We are also told that a(24) = 1.2. We use this to find k:
a(24) = 5e−^24 k^ = 1.2 =⇒ − 24 k = ln
so k = −
ln
(b) Immediately after the second dose, the patient has 6.2 units of drug in the bloodstream: 1.2 units are left over from the initial dose, and 5 more have been added by the second dose. We can now redefine time so that t = 0 is the time when the second dose was administered. Then we have a(0) = 6.2, so a(t) = 6. 2 e−kt, where k is the same as before. Then 24 hours later, we have a(24) = 6. 2 e(−^0 .05946)(24)^ ≈ 1 .488. Thus, at the end of the second 24 hours, there are 1.488 units of the drug in the bloodstream.
(c) Let ai be the amount of drug in the bloodstream at the end of day i, just before the next dose of 5 units is given. We were told a 1 = 1.2, and in (b) we computed a 2 = 1.488. To compute a 3 , a 4 , etc., we follow the same procedure that we used in (b):
a 3 = (5 + 1.488)e−(0.05946)(24)^ = 1. 557 , a 4 = (5 + 1.557)e−(0.05946)(24)^ = 1. 574 ,
and we see that the pattern is ai = (5 + ai− 1 )e−(0.05946)(24). Continuing with the calculations, we find
a 5 = (5 + 1.574)e−(0.05946)(24)^ = 1. 578 , a 6 = (5 + 1.578)e−(0.05946)(24)^ = 1. 579 , a 7 = (5 + 1.579)e−(0.05946)(24)^ = 1. 579.
Thus, when we keep only four significant digits in our calculations, the amount of the drug in the bloodstream just before each dose converges to the value 1.579. If we kept more digits, the value would continue to increase, but it would approach a limiting value. We can check this by using the equation
for ai given above. We look for a “fixed point” of this equation; that is, a value where ai = ai− 1. If we call this value a∗, the equation becomes
a∗^ = (5 + a∗)e−(0.05946)(24)
and solving for a∗^ gives a∗^ = 1. 579106.. ..
In terms of the original problem, (5 + a∗) is precisely the amount of the drug for which the amount “cleared” during a 24 hour period is 5 units.
The following graph shows the amount of the drug in the bloodstream during the first four days. The dotted line shows a∗^ = 1.579106.
0 24 48 72 96 0
1
2
3
4
5
6
7
a = a*
t (hours)
a
Amount of Drug in the Bloodstream
Note: It is possible to find a formula for ai explicity in terms of i, rather than as the recursion relation given on the previous page. If r = e(−^0 .05946)(24)^ ≈ 0 .2400, then
ai = 5r
1 − ri 1 − r
See me if you would like to see the derivation, but try it yourself first! (It is not difficult, but you will need the formula for the sum of a finite geometric series.) We can find a∗^ by taking the limit as i → ∞. Since 0 < r < 1, ri^ → 0 as i → ∞; thus
a∗^ = lim i→∞ ai =
5 r 1 − r
