Problem involving calculus

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.

• 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 – jimmy hope May 12 at 12:43
• 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. – G Aker May 12 at 12:46
• It sure did, i don't know how to solve the differential equation gotten, that is , Q'(t)= 1-Q/5 – jimmy hope May 12 at 13:44
• $\frac{DQ}{DT }= 1 - \frac{Q}{5} \implies \frac {DT}{DQ} = \frac{5}{5-Q}$ Can you solve this one? – G Aker May 12 at 13:57
• No I can't, can you help? To solve the differential equation of the question – jimmy hope May 12 at 16:34