A tank initially holds $10$ gallons of fresh water. At $t=0$, a brine solution containing $\frac 12$ pound of salt per gallon is poured into the tank at a rate of $2$ gal/min, while the well stirred mixture leaves the tank at the same rate.

$1)$ Find the amount and $2)$ the concentration of salt in the tank at any time, $t$

I have being able to differential equation by finding the values of rate in and rate out , my differential equation is DQ/DT + Q/5 = 1 . where Q is the amount of salt in the tank at time t


Your equation is correct. You can rearrange for $DQ/DT$ then take recipricols for $DT/DQ$ and solve the differential equation.

$\frac{DT}{DQ} = \frac{5}{5-Q} \implies t = \int_{Q=Q_0}^{q}\frac{5}{5-Q}DQ$

where $Q_0$ is the amount of salt at $t=0$ and $q$ is the amount of salt at time $t$. We get

$t = 5[-\ln(5-Q)]_{Q=Q_0}^q = -5(\ln(5-q)-\ln(5-Q_0))$.

Clearly $Q_0 = 0 $ so

$t = -5(\ln(5-q)-\ln(5))$

Now just rearrange for $q$.

Plotting this on a graph shows the amount of salt tending assymptotically towards 5 pounds which is what we expect.

  • $\begingroup$ Please check again, the formula for rate out is = concentration of salt in tank at time, t ) *(RATE OF OUTFLOW) = Q/10 *2= Q/5 $\endgroup$ – jimmy hope May 12 at 12:43
  • $\begingroup$ For the formula Q is not the concentration but the amount of salt in pounds. Concentration is therefore Q/10 in lbs/gallon and rate of outflow is 2 gallons per minute so rate of salt outflow is 2Q/10 = Q/5 lbs per minute. Hope this helps. $\endgroup$ – G Aker May 12 at 12:46
  • $\begingroup$ It sure did, i don't know how to solve the differential equation gotten, that is , Q'(t)= 1-Q/5 $\endgroup$ – jimmy hope May 12 at 13:44
  • $\begingroup$ $\frac{DQ}{DT }= 1 - \frac{Q}{5} \implies \frac {DT}{DQ} = \frac{5}{5-Q}$ Can you solve this one? $\endgroup$ – G Aker May 12 at 13:57
  • $\begingroup$ No I can't, can you help? To solve the differential equation of the question $\endgroup$ – jimmy hope May 12 at 16:34

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