Integration separation of variable Liquid is flowing into a small tank which has a leak. Initially the tank is empty and, t minutes later,
the volume of liquid in the tank is V cm3
. The liquid is flowing into the tank at a constant rate of
80 cm3
per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any
instant, is equal to kV cm3
per minute where k is a positive constant.

The equation becomes $ dv/dt=80 - kv  $ but how to separate it that dv one side and dt on other side ?
$$\frac {dv}{80-kv}=dt$$ and then carry out your work

since k is a constant you can do this
it should be :
after integration,
you will get
$t=\frac{-1}{k} \ln(80-kV)$

Let $V=0$ , $t=0$ 
obtain 2 equations, 
combine them and you will get 
$t-\frac{1}{k} \ln(80) = \frac{-1}{k} \ln(80-kV)$

convert in to exponential form, 
and hence you will get the final answer as
$V=\frac{1}{k}(80-80e^{-kt}$