SOLUTION:
We have to differentiate the equation only once to eliminate the arbitrary constant C.
[tex] (y-C)^2 = Cx \longrightarrow \text{Eq. 1} [/tex]
[tex] 2(y - C)y' = C [/tex]
[tex] y - C = \dfrac{C}{2y'} \longrightarrow \text{Eq. 2}[/tex]
Substituting to the given equation or Eq. 1,
[tex] \left(\dfrac{C}{2y'}\right)^2 = Cx [/tex]
[tex] \dfrac{C^2}{4(y')^2} = Cx [/tex]
[tex] \dfrac{C}{4(y')^2} = x [/tex]
[tex] C = 4x(y')^2 [/tex]
Now substitute to eq. 2,
[tex] y - 4x(y')^2 = \dfrac{4x(y')^2}{2y'} [/tex]
[tex] \therefore \boxed{y = 4x(y')^2 + 2xy'} [/tex]
