Integration of separate variables

[Milly]

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 why can't I use dv(gain)/dt = 80 and dv(lost)/dt = kv intergrate both equation and minus V(lost) from V(gain) to get V ?

 

[I like Serena]

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 why can't I use dv(gain)/dt = 80 and dv(lost)/dt = kv intergrate both equation and minus V(lost) from V(gain) to get V ?

Hi Milly! :)

You could, but then you need another DE since your second DE depends on v, which is different from v(lost).

You would need the additional equation dv = dv(gain) - dv(lost) to complete the set of differential equations.
Then, when you make the proper substitutions, you'll get dv/dt=80−kv.

 

[Milly]

What is DE?

 

[I like Serena]

What is DE?

DE stands for differential equation.