Applications of the Initial Value Problem y' = f(x, y); y(c) = d 159a. A graph of y(t) on [O, 1] is displayed in Figure 3.21.
b. As noted above 0 :=:; y(t) :=:; 3 and from Figure 3.21 or the corresponding
table of values for the numerical solution, we see t hat limt-+oo y(t) = 3.
c. From the table of values for the numerical solution we see that y(t) =
50% x 3 = 1.5 moles/liter when t is approximately. 06 seconds and we
also see that y(t) = 90% x 3 = 2.7 moles/liter when tis approximately
. 24 seconds.
d. As stated earlier CA(t) = 7 - y(t) and CB(t) = 3 -y(t).
e. Since limt-+oo y(t) = 3, we easily compute
and32.52y(x) 1.5
10.50lim CA(t) = CA(O) - lim y(t) = 7-3 = 4 moles/liter
t-+oo t-+oolim CB(t) = CB(O) - lim y(t) = 3 - 3 = 0 moles/liter.
t-HXJ t-HXJ0 0.2 0.4 0.6 0.8
xFigure 3. 21 Numerical Approximation to the IVP:y' = 2(7 - y)(3 - y); y(O) = 0
1