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

1 Answer
Tom · Jim H
Nov 13, 2017

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)#

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