Question

solve problem 23 under the assumption that the solution is pumped out at a faster rate of 10 gal/min when is the tank empty

Answer 1

$$\dfrac{dA}{dt}=rate\ in-rate\ out$$

$$\dfrac{dA}{dt}=R_{in}-R_{out}$$

$$R_{in}=(2\ \dfrac{lbs}{gal})(5\ \dfrac{gal}{min})=10\ \dfrac{lbs}{min}$$

$$R_{out}=(\dfrac{A}{500-5t}\dfrac{lbs}{gal})(10\ \dfrac{gal}{min})=\dfrac{2A}{100-t}\ \dfrac{lbs}{min}$$

$$\dfrac{dA}{dt}=10-\dfrac{2A}{100-t}, A(0)=0$$

$$\dfrac{dA}{dt}+(\dfrac{2}{100-t})A=10$$

$$\mu(t)=e^{\int{2 \over 100-t}dt}=e^{-2\ln(100-t)}=(100-t)^{-2}$$

$$(100-t)^{-2}\dfrac{dA}{dt}+2(100-t)^{-3}A=10(100-t)^{-2}$$

$$\dfrac{d}{dt}((100-t)^{-2}A)=10(100-t)^{-2}$$

Integrate

$$\int d((100-t)^{-2}A)=\int10(100-t)^{-2}dt$$

$$(100-t)^{-2}A=10(100-t)^{-1}+C$$

$$A=10(100-t)+C(100-t)^2$$

$$A(0)=0:0=10(100-0)+C(100-0)^2$$

$$=>C=-\dfrac{1}{10}$$

$$A(t)=10(100-t)-\dfrac{1}{10}(100-t)^2$$

$$T=\dfrac{500\ gal}{10\ \dfrac{gal}{min}-5\ \dfrac{gal}{min}}=100\ min$$

The tank will be empty after 100 min.

A First Course in Differential Equations with Modeling Applications 10th Edition

by Dennis G. Zill, page 91