Applications of the Initial Value Problem y' = f(x, y); y(c) = d 147I EXAMPLE 1 One Tank Mixture Problem
A 200-gallon capacity tank initially contains a salt solution consisting of
20 lbs of salt dissolved in 165 gallons of water. A salt solution with concen-
tration of 2 lbs/gal enters the tank at the rate of 8 gal/min and the resulting
uniform mixture leaves the tank at a rate of 3 gal/min. Compute and graph
the amount of salt in the tank as a function of time and find the amount of
salt in the tank at the time the tank starts to overflow.
SOLUTION
Let q(t) denote the number of pounds of salt in the tank at time t. Initially
the tank contains 20 lbs of salt, so q(O) = 20 lbs. The rate at which the
solution enters the tank is rin(t) = 8 gal/min and the concentration of the
entering solution is Cin(t) = 2 lbs/gal. The rate at which the solution leaves
the tank is rout(t) = 3 gal/min. So the rate at which the tank is filling is
r(t) = Tin(t) - rout(t) = 8 gal/min -3 gal/min= 5 gal/min. Let n(t) be the
number of gallons of solution in the tank at time t. Since the tank initially
contains 165 gallons and the rate of increase is 5 gal/min, n(t) = 165 + 5t.
Hence, the concentration of the salt solution in the tank and flowing out of
the tank at time tis Cout(t) = q(t)/n(t) = q(t)/(165 + 5t). So the initial value
problem which must be solved is
(2)dq- = Tin(t)Cin(t) - Tout(t)Cout(t)
dt
= (8 gal/min)(2 lbs/gal) - (3 gal/min) q(t) (lbs/gal)
165 + 5t
= 16 -^3 q(t) (lbs/min);
165 + 5t
q(O) = 20 lbs.The tank is full when n(t) = 165 + 5t = 200. Solving for t, we find
t = (200 - 165)/5 = 7 minutes.
Associating q(t) with y(x), we used SOLVEIVP to solve the initial value
problem (2) on the interval [O, 7] by setting f(x, y) = 16 - 3y/(165 + 5x),
by inputing the interval of integration as [O, 7], and by inputing the initial
condition y(O) = 20. A graph of the amount of salt in the tank is shown
in Figure 3.15. From the table of values for the numerical solution, we find
q(7) = 123.7934 lbs.