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## Homework Statement

Alright, I know this has been done in various incarnations but I haven't found an example exactly similar to mine..

A tank contains 60kg of salt and 1000L of water. Pure water enters the tank at the rate of 8L/min. The solution is thoroughly mixed and drains at a rate of 4L/min. Find the amount of salt in the tank after 2 hours.

## Homework Equations

y(t) = amount of salt in the tank after t minutes

v(t) = volume of water in the tank after t minutes

v(t) = 1000 + 8t-4t = 1000 + 4t

dy/dt = rate in - rate out

dy/dt = 8L/min * 0kg/L - y kg/(1000 + 4t L) * 4L/min

dy/dt = - 4y/(1000+4t)

## The Attempt at a Solution

Separating...

dy/4y = - dt/(1000 + 4t)

Integrating both sides...

(1/4)Ln(y) = -(1/4)Ln(250+t)+c

y^(1/4) = (e^c)/(250+4t)^(1/4)

Let k be the arbitrary constant = e^c

y^(1/4) = k/(250 + 4t)^(1/4)

y = k/(250+4t)

Using the initial conditions...

60 = k/(250 + 4*0)

k = 15000

y = 15000/(250 + 4t) <--- this is not right

y(120) = 20.5479 <--- this is not right

The correct answer = 40.5405405405...