How to write the differential equation

Question

A tank consists of $50$ litre of fresh water . Two litres of brine each litre containing $5$ gms of dissolved salt are run into tank per minute . The mixture is kept uniform by stirring and runs out at the rate of $1$ litre per minute . If $m$ grams of salt are present in the tank after $t$ minute , we have to expess $m$ in terms of $t$ .

I tried it and write it as $$ m = 10t - \frac{10t}{50 + t} $$

Answer 1

We measure the time $t$ in minutes, the volume $v$ of the liquid in litre and the amount of salt in gram.

At the beginning ($t=0$) the tank contains $50$ litre, so $$v(0)=50 \tag{1}$$ The volume of liquid in the tank after $t$ minutes is $v(t)$ litre. $\Delta t$ minutes later, at the moment $t+\Delta t$ the tank contains $$v(t+\Delta t) = v(t)+ 2\Delta t - \Delta t \tag{2}$$ litre. From this we get $$\frac{v(t+\Delta t) - v(t)}{\Delta t} = 1 \tag{3}$$ Assuming $\Delta t \to 0$ this gives $$v'(t)=1\tag{4}$$ From $(1)$ and $(4)$ we get $$v(t)=50+t\tag{5}$$ Of course one can get this result without using a differential equation.

$m(t)$ is the amount of salt (in gram) at the moment $t$ (in minutes) in the tank. We have $$m(0)=0\tag{6}$$ gram salt in the tank at $t=0$ (fresh water without salt) The concentration of salt at the moment $t$ is the amount of salt divided by the amount of liquid, so the concentration (in gram/litre) is $\frac{m(t)}{v(t)}$. From $t$ to $t+\Delta t$ we will loose $\frac{m(t)}{v(t)}\Delta t$ grams of salt (this is only an approximation) and get $2\cdot 5\Delta t$ of salt. So for small $\Delta t$ we have approximately $$m(t+\Delta t)=m(t)+2\cdot 5 \Delta t-\frac{m(t)}{v(t)} \Delta t \tag{7}$$ Dividing $(7)$ by $\Delta t$ we get $$\frac{m(t+\Delta t)-m(t)}{\Delta t}=10-\frac{m(t)}{v(t)}\tag{8}$$

Now we replace $v(t)$ by $(5)$ and let $\Delta t \to 0$ in $(7)$ and we finally get the differential equation

$$m'(t)=10-\frac{m(t)}{50+t}\tag{9}$$ with the initial condition $(6)$.

Answer 2

Let use denote $m(t)$ as the amount of salt presented in the tank at time $t$.

Observe the differential equation is given by \begin{align} \text{Rate of Change salt in tank} = \text{ Rate of salt entering the tank}-\text{ Rate of salt existing the tank} \end{align} or more mathematically we have \begin{align} \frac{dm}{dt} = 10 \text{ g}\cdot \text{min}^{-1} - (1\text{ L}\cdot\text{ min}^{-1})\times \frac{m(t)}{V(t)} \text{ g}\cdot \text{ L}^{-1}. \end{align}