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An acid solution flows at a constant rate of 6L/min into a tank which initially holds 200L of a 0.5% acid solution. The solution in the tank is kept well mixed and flows out of the tank at 8L/min. If the solution entering the tank is 20% acid then determine the volume of acid in the tank after t minutes.
I just want to make sure I have the differential equation right.
let y(t) be the amount of acid solution in the tank.
rate in = 0.2 * 6 = 1.2L acid per min
rate out = y(t)/v(t) * 8, where v(t) = 200 - 2t is the volume of liquid in the tank
So I got
[tex] \frac{dy}{dt} + \frac{8}{200-2t}y=1.2[/tex]
Can someone please confirm that this is the right equation to be working with?
I just want to make sure I have the differential equation right.
let y(t) be the amount of acid solution in the tank.
rate in = 0.2 * 6 = 1.2L acid per min
rate out = y(t)/v(t) * 8, where v(t) = 200 - 2t is the volume of liquid in the tank
So I got
[tex] \frac{dy}{dt} + \frac{8}{200-2t}y=1.2[/tex]
Can someone please confirm that this is the right equation to be working with?