A tank initially contains 50 gallons of brine, with 30 pounds of salt in solution. Water runs into the tank at 6 gallons per minute and the well-stirred solution runs out at 5 gallons per minute. How long will it be until there are 25 pounds of salt in the tank?

I set up the differential equation as $dQ/dt=\frac{6-5Q}{(50+t)}$.

Then I used the integrating factor method to ultimately come out with $Q=\frac{t+50+C}{(t+50)^5}$. When I plug in the initial value $Q(0)=30$, I get an outrageously high number for $C$, which may be correct, but it asks how long it will take until there are 25 pounds of salt left in the tank, so logically I plug in 25 to $Q$ and solve for $t$ to find the time it will take but I keep coming out with a negative number, which makes no sense.

Any guidance is helpful.


The amount change of brine $V(t)=(6-5)t=t$;

Amount of salt in equals $0$ since there is NO SALT in;

Amount of salt out equals $\frac{5Q}{V(t)}=\frac{5Q}{(50+t)}$;

Then try $dQ/dt$ = In-Out.

  • $\begingroup$ can you explain that again? How did you get the amount of salt out equation? $\endgroup$ – user333673 Jun 30 '16 at 5:43
  • $\begingroup$ Q/V is the density of the salt at a certain moment, and there are 5 gallons solution running out. Therefore, the total amount of salt running out will be 5Q/V. Hope it helps. $\endgroup$ – Shizhuo Jun 30 '16 at 5:46
  • $\begingroup$ Good answer! I edited your answer to have LaTeX syntax. I know you're brand new here. Welcome to Math SE first of all! Secondly, I suggest learning LaTeX for Math SE. If you take a look around, you can see that mostly everyone here uses it. It makes equations look really pretty. $\endgroup$ – KingDuken Jun 30 '16 at 14:00
  • $\begingroup$ Thank you so much! I will try to learn LaTex. $\endgroup$ – Shizhuo Jun 30 '16 at 18:04

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