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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?