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$V=$ volume of water in the water tank

$M(t) =$ mass of sands in the water at time $t$

$K(t) =$ Concentration of sands in water at time $t$

$R=$ rate of water flowing out

Concentration sands in water coming in is constant.

If $K(t)$ is the concentration of sands in water flowing out, and $K_{in}$ the concentration flowing in. Then i have this equation $$\frac{dK}{dt}=\frac{R}{V}K_{in} - \frac{R}{V}K$$

How do i solve this differential equation?

If $K_{in} = 10$, what is concentration of sands in the water in long run?

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  • $\begingroup$ Do you know about $e^t$? $\endgroup$ – Michael Galuza Aug 3 '15 at 7:51
  • $\begingroup$ @MichaelGaluza are you talking about properties of Lapace transform? i don't know how i can relate. if not , i don't konw what about et $\endgroup$ – problematic Aug 3 '15 at 7:54
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Notice, $$\frac{dK}{dt}=\frac{R}{V}K_{in}-\frac{R}{V}K$$ $$\implies \frac{dK}{dt}=\frac{R}{V}(K_{in}-K)$$ $$ \frac{dK}{K-(K_{in})}=-\frac{R}{V}dt$$ Since, $K_{in}$, concentration sands flowing in, is constant hence integrating both the sides w.r.t. time $t$, we get $$ \int\frac{dK}{(K-K_{in})}=-\int \frac{R}{V}dt$$ $$\ln(K-K_{in})=-\frac{R}{V}t+c$$ $$\ln(K-K_{in})-\ln C=-\frac{R}{V}t$$ $$\ln\left(\frac{K-K_{in}}{C}\right)=-\frac{R}{V}t$$ $$\implies K-K_{in}=Ce^{\frac{-R}{V}t}$$ $$\implies K=K_{in}+Ce^{\frac{-R}{V}t}$$

Where $C$ is a constant. Now taking the limit at $t\to \infty$ as follows $$$$ the concentration sands in water in long run is given as follows $$\lim_{t\to \infty}K=\lim_{t\to \infty}\left(K_{in}+Ce^{\frac{-R}{V}t}\right)$$ $$K_{\infty}=K_{in}+C(0)$$$$=K_{in}=10$$

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  • $\begingroup$ Kindly, go through the steps in my answer giving proof of that $\endgroup$ – Harish Chandra Rajpoot Aug 3 '15 at 15:08
  • $\begingroup$ Yes, you will get the same $K$ in long run by taking limit at $t\to \infty$ $\endgroup$ – Harish Chandra Rajpoot Aug 3 '15 at 16:59
  • $\begingroup$ i can't, i will just use your solution $\endgroup$ – problematic Aug 3 '15 at 18:55
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Substitution: $K_{1} = K - K_{in} $. Then equation is $$\frac{d K_1}{d t} = -\frac{R}{V}K_1$$ Solution is $ K_1(t) = C * e^{-\frac{R}{V} t}$ where C is constant difined from boundaries i.e. from t=0. Then $$ K = K_{in} + C * e^{-\frac{R}{V} t} $$ In long run $ K = K_{in} $.

NB: Solution assumes that $ K_{in} $ is constant and doesn't depend from t.

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