Applications of the Initial Value Problem y' = f(x, y); y(c) = d 149
dq1
----;Ji"= run(t)cLin(t) - ri_out(t)CLout(t)= (5 gal/min)(.1) - (5 gal/min) (i~^1 0)
= (.5 - ~~) (gal/min).Tank I
150 galr 1 ou/t) = 5 gal/min
cl~ozll(t)r 2 in (t) = 5 gal/min
c2=in (t)Tank2
75 galFigure 3.16 Diagram for a Two Tank Mixture ProblemFor tank 2, we have r2_in(t) = r2_out(t) = 5 gal/min, c2_in(t) = Qi(t)/150
and C2_out(t) = Q2(t)/75. So the number of gallons of dye in tank 2 must
satisfy the differential equation
dq2(t)
~ = r2_in(t)c2_in(t) - r2_out(t)c2_out(t)= (5 gal/min) ( q;) -(5 gal/min) ( q))
= (qi(t) - q^2 (t)) (gal/min).
30 15Hence, the number of gallons of dye in tanks 1 and 2 must simultaneously
satisfy the two initial value problems
(3a)(3b)dq1 - 5 - :l!._.
dt -. 30'dq2 Q1 Q2
dt 30 - 15 ;