A tank initially contains 60 gal of pure water. Brine containing 1 lb of salt per gallon enters the tank at 2 gal/min, and the (perfectly mixed) solution leaves the tank at 3 gal/min; thus the tank is empty after exactly 1 h.

(a) Find the amount of salt in the tank after t minutes.

(b) What is the maximum amount of salt ever in the tank?

My working:

Let x(t) (in lb) be the amount of salt in the tank at time t.

(a) $\displaystyle \frac{dx}{dt}=r_ic_i-r_oc_o=2*1-3*\frac{x}{60-t}=2-\frac{3x}{60-t}$

$\displaystyle \frac{dx}{dt}+\frac{3x}{60-t}=2$

$\displaystyle \rho(t)=exp\left(\int \frac{3}{60-t}dt\right)=exp(-3ln\ (60-t))=\frac{1}{(60-t)^3}$

Multiplying by $\displaystyle \rho(t)$, we get

$\displaystyle \frac{1}{(60-t)^3}\frac{dx}{dt}-\frac{3x}{(60-t)^4}=\frac{2}{(60-t)^3}$

$\displaystyle \frac{d}{dt}\left[\frac{x}{(60-t)^3}\right]=\frac{2}{(60-t)^3}$

$\displaystyle \frac{x}{(60-t)^3}=2\frac{(60-t)^{-2}}{2}+C$

$\displaystyle \frac{x}{(60-t)^3}=\frac{1}{(60-t)^2}+C$

$\displaystyle x=(60-t)+C(60-t)^3$

$\displaystyle x(0)=0$

$\displaystyle 0=60+C*60^3$

$\displaystyle C=-\frac{1}{3600}$

$\displaystyle x=(60-t)-\frac{(60-t)^3}{3600}$

Solved it! Don't bother.