# Mixing Problem: gallons in tank at a time t

I have this problem

I know that we can write the initial value problem as

$$\frac{dy}{dt} = 5 - \frac{3y}{100+2t},\quad y(0)=50$$ but how do i solve this?

i was doing...
$$p(t)=e^{\large\int \frac{3}{100+2t}dt}=e^{\frac{3}{2}\ln(100+2t)}$$

but what else should be doing?

also do you know a site where I can read more examples about cisterns?

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Use that function to factor the differential equation: $(py)' = 5p$, integrate both sides with a $+C$, solve for the constant using the initial conditions, then isolate $y$ on one side to fully solve for it. You might want to google "mixing problems differential equations" or some such phrase for more information. – anon Jul 14 '11 at 3:14
@Zev: good edit, I had read "galois in tank at a time $t$" which was slightly disturbing! :D – lentic catachresis Jul 14 '11 at 4:01
@Bruno: Thanks! I made that same misread :) – Zev Chonoles Jul 14 '11 at 4:06

For more, search for "Mixing Problems" or first order differential equations. There are many examples in Calculus in One and Several Variables by Salas, Hille, and Etgen - and essentially every ODE textbook will start with equations of about this calibre.

Now, to solve this. You have $e^{\frac{3}{2} \ln {(100 + 2t)} dt}$. Of course, in general $e^{\ln {g} } = g$, as log and e are inverse functions. So this simplifies very nicely. I suspect you did this because when you multiply by this "integrating factor", you have an equation of the form $\frac{d}{dt} (p(t) y(t) )$. So then you can integrate out, add a constant, and write out the answer.

Is that a good enough hint?

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ok, I am following that hint! – cMinor Jul 14 '11 at 3:22