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Suppose that a large mixing tank initially holds 500 gallons of water in which 60 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 5 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0. What is A(0)

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Answer:

The answer is "[tex]\frac{dA}{dt}+ \frac{A}{100}=0\ \ and \ \A(0)=60[/tex]"

Step-by-step explanation:

[tex]A(0)=60\\\\ \frac{dA}{dt}=Rin-Rout\\\\Rin=0\\\\Rout=\frac{A(t)}{500} \times (5)\\\\ =\frac{A}{100}\\\\ so,\\\\\frac{dA}{dt}=Rin-Rout\\\\\frac{dA}{dt}=0- \frac{A}{100}\\\\\frac{dA}{dt}+ \frac{A}{100}=0\\\\A(0)=60[/tex]

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