# Mixing Problem (variable volume)

## Homework Statement

A tank contains 150L of water in which initially 30g or salt is dissolved. Brine runs in at a rate of 4 L/m and each litre contains 1.5g of dissolved salt. The mix in the tank is kept uniform. Brine runs out at 3L/m. What is that amount of salt in the tank at any time t?
Note that the volume of brime in the tank is not constant in time

## Homework Equations

$$y' +p(x)y = r(x)$$

General solution to first order, linear ODE

$$y(t) = exp(-\intp(p(t)) dt) [\int r(t)\exp(\intp(p(t))dt) dt +C]$$

## The Attempt at a Solution

Input/Output Gives

$$y(t)=(6-3y)/(150+t)$$
-
$$y'+3y = 6/(150+t)$$

Substituting these values into the general equation gives me

y(t) = e^(-3t)[\nt(6/(150+t)*e^(3t) dt) + C]

And from here ... well... some sort of a clue as to where to go next would be very nice.
Basically i am having trouble with that integral, is it possible to take $$\int(ln((6/(150+t)*\exp^(3t) dt)$$ somehow? I'm not sure if/how to make this step.
I can take the log of the other parts of the equation...

ln(y(t)) = ln(exp^3t)+ [this is the bit i'm not sure of] + ln(C)

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