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.

The amount of salt in the tank changes with rate according to[tex]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)[/tex][tex]\implies Q'+\dfrac Q{300+3t}=4[/tex]which is a linear ODE in [tex]Q(t)[/tex]. Multiplying both sides by [tex](300+3t)^{1/3}[/tex] gives[tex](300+3t)^{1/3}Q'+(300+3t)^{-2/3}Q=4(300+3t)^{1/3}[/tex]so that the left side condenses into the derivative of a product,[tex]\big((300+3t)^{1/3}Q\big)'=4(300+3t)^{1/3}[/tex]Integrate both sides and solve for [tex]Q(t)[/tex] to get[tex](300+3t)^{1/3}Q=(300+3t)^{4/3}+C[/tex][tex]\implies Q(t)=300+3t+C(300+3t)^{-1/3}[/tex]Given that [tex]Q(0)=100[/tex], we find[tex]100=300+C\cdot300^{-1/3}\implies C=-200\cdot300^{1/3}[/tex]and we get the particular solution[tex]Q(t)=300+3t-200\cdot300^{1/3}(300+3t)^{-1/3}[/tex][tex]\boxed{Q(t)=300+3t-2\cdot100^{4/3}(100+t)^{-1/3}}[/tex]