Answer to Question #267087 in Differential Equations for mary

Question

1.   A tank initially contains 50 gal of fresh water. At t=0, brine solution containing 2 lb of salt per gallon is poured into the tank at the rate of 12 gal/min, while the well stirred mixture leaves the tank at the rate of 8 gal/min. what is the amount of salt at the end of 5 minutes? How long will it take to obtain an amount of 50 lb? ans. 88.98 lb; 2.47 min
2. Given an RC series circuit that has an emf source of 50 volts, a resistance of 20 kilo ohms, a capacitance of 6 microfarad and the initial charge of the capacitor of the capacitor is 1 micro coulomb. What is the charge in the capacitor at the end of 0.01 second? What is the current in the circuit at the end of 0.05 seconds? Ans. 24.91µC; 1.64mA

Answer 1

1. Let $s(t) =$ amount, in lb of salt at time $t.$ Then we have

$\dfrac{ds}{dt}=$(rate of salt into tank) − (rate of salt out of tank)

$$\dfrac{ds}{dt}=2\cdot12-\dfrac{8s}{50+(12-8)t}$$

So we get the differential equation

$$\dfrac{ds}{dt}=24-\dfrac{8s}{50+4t}$$

$$\dfrac{ds}{dt}+\dfrac{4s}{25+2t}=24$$

Integrating factor

$$\mu(t)=e^{\int 4dt/(25+2t)}=(25+2t)^{2}$$

$$\dfrac{d}{dt}((25+2t)^{2}s)=24(25+2t)^{2}$$

$$\int d((25+2t)^{2}s)=24\int(25+2t)^{2}dt$$

$$(25+2t)^{2}s=24(\dfrac{1}{6})(25+2t)^{3}+c_1$$

$$s(t)=4(25+2t)+c_1(25+2t)^{-2}$$

$$s(0)=100+c_1(25)^{-2}=0$$

$$c_1=-62500$$

$$s(t)=4(25+2t)-62500(25+2t)^{-2}$$

$$s(5)=4(25+2(5))-62500(25+2(5))^{-2}$$

$$s(5)=88.98\ lb$$

$$s(t)=4(25+2t)-62500(25+2t)^{-2}=50$$

Let $y(t)=s(t)-50$

$$y(t)=4(25+2t)-62500(25+2t)^{-2}-50$$

Solve graphically the equation $y=0$

$t=2.47\ min$

2.

$$Rq'+\dfrac{1}{C}q=V$$

$$2\cdot10^{4}q'+\dfrac{1}{6\cdot10^{-6}}q=50$$

$$q'+\dfrac{25}{3}q=0.0025$$

$$q'=-\dfrac{25}{3}(q-0.0006)$$

$$\dfrac{dq}{q-0.0003}=-\dfrac{25}{3}dt$$

Integrate

$$q=0.0003+c_1e^{-25t/3}$$

Given $q(0)=10^{-6}$

$$0.000001=0.0003+c_1$$

$$c_1=-0.000299$$

$$q(t)=0.0003-0.000299e^{-25t/3}$$

$$q(0.01)=0.0003-0.000299e^{-25(0.01)/3}$$

$$=24.91\cdot10^{-6} (C)$$

$24.91\ \mu C$

$$i(t)=q'(t)=-0.000299(-\dfrac{25}{3})e^{-25t/3}$$

$$=\dfrac{0.007475}{3}e^{-25t/3}$$

$$i(0.05)=\dfrac{0.007475}{3}e^{-25(0.05)/3}=0.0016426(A)$$

$$=1.64(\ mA)$$

$1.64\ mA$