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
[tex] y' +p(x)y = r(x) [/tex]
General solution to first order, linear ODE
[tex] y(t) = exp(-\intp(p(t)) dt) [\int r(t)\exp(\intp(p(t))dt) dt +C] [/tex]
The Attempt at a Solution
[tex]y'+3y = 6/(150+t)[/tex]
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 [tex]\int(ln((6/(150+t)*\exp^(3t) dt)[/tex] 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)