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.
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
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...