Question

Suppose that a large mixing tank initially holds 700 gallons of water in which 50 pounds of salt have been dissolved. another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. if the concentration of the solution entering is 4 lb/gal, determine a differential equation for the amount of salt a(t) in the tank at time t > 0. (use a for a(t).)

Answer 1

$\dfrac{\mathrm dA}{\mathrm dt}=\dfrac{3\text{ gal}}{1\text{ min}}\dfrac{4\text{ lb}}{1\text{ gal}}-\dfrac{2\text{ gal}}{1\text{ min}}\dfrac{A\text{ lb}}{700+(3-2)t\text{ gal}}$
$\iff A'+\dfrac2{700+t}A=12$

$(700+t)^2A'+2(700+t)A=12(700+t)^2$
$((700+t)^2A)'=12(700+t)^2$
$(700+t)^2A=4(700+t)^3+C$
$A=4(700+t)+\dfrac C{(700+t)^2}$

Initially, the tank contains 50 lbs of salt in 700 gal of solution, i.e. $A(0)=\dfrac{50}{700}=\dfrac5{70}$, so

$\dfrac5{70}=2800+\dfrac C{700^2}$

Use this to solve for $C$.