Applications of the Initial Value Problem y' = f(x, y); y(c) = d 151
Associating q2(t) with y(x), we used SOLVEIVP to solve the initial value
problem (5) on [O, 125] by setting f(x, y) = 2.5e-x/^3 o + .5 -y/ 15 , by inputing
the interval of integration as [O, 125], and by entering the initial condition
y(O) = 0. A graph of the solution for the number of gallons of dye in tank 2
on the interval [O, 125] is shown in Figure 3.17. From the graph we see that the
maximum number of gallons of dye in the tank occurs when t is approximately
25 minutes. By searching the table of values of the numerical solution near
t = 25 minutes, we find the maximum number of gallons of dye in the tank is
24 .545 gallons and that the maximum occured when t = 23.625 minutes.
25
20
15
y(x)
10
5
0
0 20 40 60 80 100
x
Figure 3.17 Numerical Approximation to the IVP:
y' = 2.5e-x/^3 o + .5 - y/15; y(O) = 0
EXERCISES 3.6
120
l. The bloodstream carries a drug to an organ at the rate of 1.5 in^3 /sec
and leaves the organ at the same rate. The organ contains 8 in^3 of
blood and the concentration of the drug in the blood entering the organ
is 0 .5 g/in^3. If at time t = 0 there is no trace of the drug in the organ,
and if we can assume that the organ behaves like a container in which
the blood is uniformly mixed, compute and graph the amount of drug
in the organ over the time interval [O, 10]. When is the amount of drug
in the organ 2 g?
- An office 12 ft by 10 ft and 8 ft high (the size of some mathematics
professors' offices) initially contains air with no carbon monoxide. Be-