Answer to Question #267087 in Differential Equations for mary

Question #267087
  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
1
Expert's answer
2021-11-17T15:24:16-0500

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

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


dsdt=2128s50+(128)t\dfrac{ds}{dt}=2\cdot12-\dfrac{8s}{50+(12-8)t}

So we get the differential equation


dsdt=248s50+4t\dfrac{ds}{dt}=24-\dfrac{8s}{50+4t}

dsdt+4s25+2t=24\dfrac{ds}{dt}+\dfrac{4s}{25+2t}=24

Integrating factor


μ(t)=e4dt/(25+2t)=(25+2t)2\mu(t)=e^{\int 4dt/(25+2t)}=(25+2t)^{2}

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

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

(25+2t)2s=24(16)(25+2t)3+c1(25+2t)^{2}s=24(\dfrac{1}{6})(25+2t)^{3}+c_1

s(t)=4(25+2t)+c1(25+2t)2s(t)=4(25+2t)+c_1(25+2t)^{-2}

s(0)=100+c1(25)2=0s(0)=100+c_1(25)^{-2}=0

c1=62500c_1=-62500

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

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

s(5)=88.98 lbs(5)=88.98\ lb


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

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


y(t)=4(25+2t)62500(25+2t)250y(t)=4(25+2t)-62500(25+2t)^{-2}-50

Solve graphically the equation y=0y=0


t=2.47 mint=2.47\ min


2.


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

2104q+16106q=502\cdot10^{4}q'+\dfrac{1}{6\cdot10^{-6}}q=50

q+253q=0.0025q'+\dfrac{25}{3}q=0.0025

q=253(q0.0006)q'=-\dfrac{25}{3}(q-0.0006)

dqq0.0003=253dt\dfrac{dq}{q-0.0003}=-\dfrac{25}{3}dt

Integrate


q=0.0003+c1e25t/3q=0.0003+c_1e^{-25t/3}

Given q(0)=106q(0)=10^{-6}


0.000001=0.0003+c10.000001=0.0003+c_1

c1=0.000299c_1=-0.000299

q(t)=0.00030.000299e25t/3q(t)=0.0003-0.000299e^{-25t/3}

q(0.01)=0.00030.000299e25(0.01)/3q(0.01)=0.0003-0.000299e^{-25(0.01)/3}

=24.91106(C)=24.91\cdot10^{-6} (C)

24.91 μC24.91\ \mu C


i(t)=q(t)=0.000299(253)e25t/3i(t)=q'(t)=-0.000299(-\dfrac{25}{3})e^{-25t/3}

=0.0074753e25t/3=\dfrac{0.007475}{3}e^{-25t/3}

i(0.05)=0.0074753e25(0.05)/3=0.0016426(A)i(0.05)=\dfrac{0.007475}{3}e^{-25(0.05)/3}=0.0016426(A)

=1.64( mA)=1.64(\ mA)

1.64 mA1.64\ mA



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