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**superdude** tank contains 1000L of water with 15kg of salt dissolved in it. Pure water enters tank at 10L/min and the mixture leaves at the same rate. How much salt is in the tank after t minutes?

Work:

let y(t)=mass of salt in water

y(0)=15kg

dy/dt = rate in - rate out

rate in = 0 because no salt is entering

rate out = $\displaystyle \frac{y(t)}{1000L}$

$\displaystyle \longrightarrow \frac{dy}{dt}=0-{y(t)}{1000L}\times\frac{10L}{min}=\frac{-y(t)}{100min}$

This becomes a seperatable equation so I get $\displaystyle \int\frac{-1}{y(t)}dy=\int\frac{1}{100t} dt = -ln|y| = \frac{ln|t|}{100}$ solving for y I get $\displaystyle \frac{1}{t^{\frac{1}{100}}}$

The answer given is $\displaystyle 15e^{\frac{-t}{100}}kg$