Hi everyone, first post on here In a bit of strife though and looking for a bit of guidance...

The Problem:

A 150 litre tank contains 40 litres of water and 20 litres of alcohol mixed together. Pure alcohol is run in at 5 litres/minute, and the mixture is stirred to keep it perfectly uniform. Simultaneously, the mixture is withdrawn at 3 litres/minute.

Derive a differential equation for the amount of alcohol, $\displaystyle x(t)$ litres, in the tank at time $\displaystyle t$ minutes.

My attempt so far:

This is my first time dealing with two liquids in a mixture (L/L) as opposed to a solid (such as a pollutant in a solution which is measured in g/L)

Let $\displaystyle x$ be the amount of alcohol in the tank at time $\displaystyle t$.

$\displaystyle \frac{dx}{dt}$ = rate alcohol flows in - rate alcohol flows out

Rate in= Concentration X Flow rate in = 1 L/L X 5 $\displaystyle \frac{L}{min}$

This is where I've hit a stumbling block as I'm having trouble calculating what rate out comes to.

Rate out= Concentration X Flow rate out = x/(60+2t) X 3 $\displaystyle \frac{L}{min}$

The reason I've gone with $\displaystyle (60 + 2t)$ as the volume is that there's 60L at the start, and the difference between the two rates of flow is $\displaystyle 2t$.Am I right in doing so?

And then the finaldifferential equationI get to is:

$\displaystyle \frac{dx}{dt}$ = 5 - x/(20 + t) (simplified from above)

A premeditated thanks for the help!