Question

How do you demonstrate that, if `(df)/(dr)= (dg)/(dy)`, then `(df)/(dr)= (dg)/(dy) = K` where K is a constant ?

Answer 1

EDIT : I'm sorry i allow myself to edit your answer and answer my own question

Explanation:

explanation here :

Let's say that if `r` varies `(df)/(dr)` varies too so the equality must be respected so it mean that `(dg)/dy` need to change too, but it doesn't depend of `r` so there is a problem.

the only way the equality is respected is that the functions are equal to the same constant.

EDIT 2:

For the math

`(df)/(dr) = (dg)/(dy)`

we derivate both side by `y`

`d/(dy)((df)/(dr)) = (d^2g)/(dy^2)`

`(d^2g)/(dy^2) = 0`

`(dg)/(dy) = K = (df)/(dr)`