Q:

Solving differential equation , Calculate concentration of sands in water?

$V=$ 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?

A:

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)$$$$=K_{in}=10$$  

