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


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  • $\begingroup$ What have you tried so far? $\endgroup$ – Henry May 12 at 7:52
  • $\begingroup$ I have done quite a few steps, but I'm not able to upland pictures. $\endgroup$ – jimmy hope May 12 at 7:53
  • $\begingroup$ I have calculated the rate in and rate out , the I got the differentiation equation. DQ/DT + Q/5 = 1 $\endgroup$ – jimmy hope May 12 at 7:54
  • $\begingroup$ What is $Q$ there? $\endgroup$ – Henry May 12 at 8:02
  • $\begingroup$ Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. $\endgroup$ – N. F. Taussig May 12 at 9:24

Call Q the amount of salt in the tank, then $Q'(t)=\text{difference of the amount of salt going in and out}$. Clearly, 1 pound of salt goes in per minute and $\frac{Q}{5}$ pounds of salt goes out.

Which gives the following separable ODE: $Q'(t)=1-\frac{Q}{5}$; $\hspace{2pt}$$\frac{Q'(t)}{1-\frac{Q}{5}}=1$

Solving it yields: $Q(t) = c_1\cdot e^{\frac{-t}{5}} + 5$ Since $Q(0)=0$, the final solution is $Q(t) = -5 \cdot e^{\frac{-t}{5}} + 5$ And $C(t)=\frac{Q(t)}{10}=\frac{-1}{2} \cdot e^{\frac{-t}{5}} + \frac{1}{2}$


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