Question

A tank holds 300 gallons of water and 100 pounds of salt. A saline solution with concentration 1 lb salt/gal is added at a rate of 4 gal/min. Simultaneously, the tank is emptying at a rate of 1 gal/min. Find the specific solution Q(t) for the quantity of salt in the tank at a given time t.

Answer 1

The amount of salt in the tank changes with rate according to

$Q'(t)=\left(1\dfrac{\rm lb}{\rm gal}\right)\left(4\dfrac{\rm gal}{\rm min}\right)-\left(\dfrac{Q(t)}{300+(4-1)t}\dfrac{\rm lb}{\rm gal}\right)\left(1\dfrac{\rm gal}{\rm min}\right)$

$\implies Q'+\dfrac Q{300+3t}=4$

which is a linear ODE in $Q(t)$. Multiplying both sides by $(300+3t)^{1/3}$ gives

$(300+3t)^{1/3}Q'+(300+3t)^{-2/3}Q=4(300+3t)^{1/3}$

so that the left side condenses into the derivative of a product,

$\big((300+3t)^{1/3}Q\big)'=4(300+3t)^{1/3}$

Integrate both sides and solve for $Q(t)$ to get

$(300+3t)^{1/3}Q=(300+3t)^{4/3}+C$

$\implies Q(t)=300+3t+C(300+3t)^{-1/3}$

Given that $Q(0)=100$, we find

$100=300+C\cdot300^{-1/3}\implies C=-200\cdot300^{1/3}$

and we get the particular solution

$Q(t)=300+3t-200\cdot300^{1/3}(300+3t)^{-1/3}$

$\boxed{Q(t)=300+3t-2\cdot100^{4/3}(100+t)^{-1/3}}$