1. Let s(t)= amount, in lb of salt at time t. Then we have
dtds=(rate of salt into tank) − (rate of salt out of tank)
dtds=2⋅12−50+(12−8)t8s So we get the differential equation
dtds=24−50+4t8s
dtds+25+2t4s=24 Integrating factor
μ(t)=e∫4dt/(25+2t)=(25+2t)2
dtd((25+2t)2s)=24(25+2t)2
∫d((25+2t)2s)=24∫(25+2t)2dt
(25+2t)2s=24(61)(25+2t)3+c1
s(t)=4(25+2t)+c1(25+2t)−2
s(0)=100+c1(25)−2=0
c1=−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′+C1q=V
2⋅104q′+6⋅10−61q=50
q′+325q=0.0025
q′=−325(q−0.0006)
q−0.0003dq=−325dt Integrate
q=0.0003+c1e−25t/3 Given q(0)=10−6
0.000001=0.0003+c1
c1=−0.000299
q(t)=0.0003−0.000299e−25t/3
q(0.01)=0.0003−0.000299e−25(0.01)/3
=24.91⋅10−6(C) 24.91 μC
i(t)=q′(t)=−0.000299(−325)e−25t/3
=30.007475e−25t/3
i(0.05)=30.007475e−25(0.05)/3=0.0016426(A)
=1.64( mA) 1.64 mA
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