(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 500-gallon tank originally contains 350 gallons of pure water. A 2-pound per gallon brine solution is pumped into the tank at 10 gallons per minute and the well-stirred mixture is pumped out at one-half that rate. How much salt is in the tank at th emoment of overflow?

Answer: 510 lb

2. Relevant equations

dA/dt = [(rate in)(Concentration in)] - [(rate out)(concentration out)]

3. The attempt at a solution

Here is what I did.

dA/dt = (2lb/gal * 10 gal/min) - [A/(350+5t) * 5]

dA/dt = 20 - A/(70+t)

Rearrange:

dA/dt + A/(70+t) = 20

Initial condition: A (0) = 0

e^{intgral[1/(70+t)]}= 70 + t

A(t) = [1/(70+t)] * [10* integral(t+70)dt + c

A(t) = 5(t+70) + c/(t+70)

A(0) = 0;

c = -24,500

Thus; A(t) = 5(t+70) - 24500/(t+70)

A (10)/500 = 400 - 24500/80

= 93.75 lb/gal

This answer is incorrect, as the answer should be 510 lb.

PLEASE HELP!!

Thanks!!

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# Homework Help: Differential Equations: Linear Models

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