First order linear differential equations The question is:
A tank with a capacity of 4000 L is full of a mixture of water and chlorine with a concentration of 0.005 g of chlorine per liter. In order to reduce the concentration of chlorine, fresh water is pumped into the tank at a rate of 40 L/s. The mixture is kept stirred and is pumped out at a rate of 100 L/s. Find the amount of chlorine in the tank as a function of time. (Let y be the amount of chlorine in grams and t be the time in seconds.)
i mainly need help with how to approach this question... i have never seen a text question which i have to understand the equation from it. i've looked up google for similar questions but it only had the final solution with no explanations...
i'd really appreciate help with how to get the equations from the text and how do i know where to start...
 A: The amount of liquid in the tank  at time $t$ is $4000-60t$. (We are using $t=0$ for the time the process begins.)  This is because liquid is entering at $40$ litres per second and leaving at $100$ litres per second. 
If $y$ is the amount of chlorine in the tank at time $t$, then the concentration of chlorine at time $t$ is $\frac{y}{4000-60t}$ grams per litre.
Since liquid is leaving the tank at $100$ litres per second, the rate at which chlorine is leaving at time $t$ is $\frac{100y}{4000-60t}$ (grams per second).
We thus obtain the differential equation
$$\frac{dy}{dt}=-\frac{100y}{4000-60t}.$$
This is a separable DE. Use standard techniques to find the solution. The initial condition is $y(0)=(4000)(0.005)$. 
Added: the calculation
The DE can be rewritten as 
$$\frac{dy}{y}=-\frac{5\,dt}{200-3t}.$$
Integrate. We get
$$\ln|y|=\frac{5}{3}\ln(|200-3t|)+C.$$
Exponentiate. We get
$$y=K(200-3t)^{5/3}.$$
Using the initial condition $y(0)=20$, we get
$$y=20(200)^{-5/3}(200-3t)^{5/3}.$$
This looks much more attractive as
$$y=20\left(1-\frac{3t}{200}    \right)^{5/3}.$$
Remark: The differential equation has a limited range of validity, since pretty soon the tank will be empty. 
