Answer to Question #273945 in Differential Equations for Peter

Question #273945

A 50 gallons tank initially contains 10 gal of fresh water. At t = 0, a brine solution


containing 1 lb of salt per gallon is poured into the tank at the rate of 4 gal/min, while the


well-stirred mixture leaves the tank at the rate of 2 gal/min. Find


a. the amount of time required for overflow to occur


b. the amount of salt in the tank at the moment of overflow

1
Expert's answer
2021-12-02T16:18:53-0500

2.

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=1(4)2s10+(42)t\dfrac{ds}{dt}=1(4)-\dfrac{2s}{10+(4-2)t}

So we get the differential equation


dsdt+s5+t=4,s(0)=0\dfrac{ds}{dt}+\dfrac{s}{5+t}=4, s(0)=0


Integrating factor


μ=5+t\mu=5+t

(5+t)dsdt+s=4(5+t)(5+t)\dfrac{ds}{dt}+s=4(5+t)

d((5+t)s)=4(5+t)dtd((5+t)s)=4(5+t)dt

Integrate


d((5+t)s)=4(5+t)dt\int d((5+t)s)=\int4(5+t)dt

(5+t)s=20t+2t2+C(5+t)s=20t+2t^2+C

s=20t+2t2+C5+ts=\dfrac{20t+2t^2+C}{5+t}

s(0)=C5=0=>C=0s(0)=\dfrac{C}{5}=0=>C=0

s(t)=20t+2t25+ts(t)=\dfrac{20t+2t^2}{5+t}

a. V=10+2t=50=>t=20V=10+2t=50=>t=20

The tank overflows at t=20t=20 min.


b.

s(20)=20(20)+2(20)25+20=48(lb)s(20)=\dfrac{20(20)+2(20)^2}{5+20}=48(lb)


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