# Prove that $$x^2-y^2\;=\;c(x^2+y^2)^2$$is the general solution of differential equation $$(x^3-3x\;y^2)dx\;=\;(y^3-3x^2y)dy$$ where $$c$$ is a parameter.

Toolbox:
• A differential equation of the form $\large\frac{dy}{dx} =$$F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. To solve such equations, substitute y = vx and \large\frac{dy}{dx} = v+$$x\large\frac{dv}{dx}.$
Step 1:
Given :$(x^3 - 3xy^2) = (y^3 - 3x^2y)dy$
Rearranging the equation we get,
$\large\frac{dy}{dx} = \frac{(x^3-3xy^2)}{(y^3 - 3x^2y)}$
This is homogenous equation of degree zero.
Hence to solve this let us substitute $y = vx$ and $\large\frac{dy}{dx} = v+$$x\large\frac{dv}{dx}. v+ x\large\frac{dv}{dx} =\frac{ (x^3 -3xv^2x^2)}{(v^3x^3 -3x^2vx)} Taking x^3 as the common factor and cancelling we get v+x\large\frac{dv}{dx} =\frac{ (1-3v^2)}{(v^3 - 3v)} Bringing v from the LHS to the RHS we get, x\large\frac{dv}{dx }=\frac{ (1-3v^2)}{v(v^2-3)}$$ - v$
$x\large\frac{dv}{dx }= \frac{(1-3v^2 -v^4 + 3v^2)}{v(v^2 - 3)}$
$x\large\frac{dv}{dx }= \frac{(1-v^4)}{v(v^2-3)}$
Seperating the variables we get,
$\large\frac{(v^3-3v)dv}{(1-v^4) }=\frac{ dx}{x}$
$\large\frac{v^3}{(1-v^4)} - \frac{3v}{(1-v^4)} = \frac{dx}{x}$
Step 2:
Integrating on both sides we get,
$\int\large\frac{v^3}{(1-v^4)} -\int \frac{3v}{(1-v^4)} =\int \frac{dx}{x}$
$\int\large\frac{ v^3}{(1-v^4) } = I_1$
This can be done by substitution method
Let $1-v^4 = t$; then $dt = - 3v^3dv$ or $dv = - dt/3$
$I_1 = - (1/3)\int dt/t =- (1/3)\log t$
Substituting for t we get
$I_1 = - (1/3)\log(1-v^4)$
Step 3:
Let $I_2$ be $\large\frac{3v}{(1-v^4)}$
This can be written as
$I_2 = 3 \int \large\frac{v}{(1-v^4)}$
$\quad=3\int\large\frac{v}{[1-(v^2)^2]}$
Let $v^2=u$
Hence $du=2vdv$
Or $dv=\large\frac{du}{2}$
Substituting this in $I_2$ we get,
$I_2=3\int\large\frac{-1}{2[du/1-u^2]}$
$I_2=\large\frac{1}{2}$$\times 2\log[\large\frac{|1-u|}{|1+u|}] \quad=3/4 \log\large\frac{|1-v^2|}{|1+v^2|} I_1 - I_2 = \large\frac{-1}{4}$$(\log(1-v^4) +\large\frac{3}{4}$$\log\large\frac{|1+v^2|}{|1-v^2|} Step 4: Substituting for v =\large\frac{ y}{x} -\large\frac{1}{4}$$\log(1-y^4/x^4) +\large\frac{ 3}{4}$$\log\large\frac{(1+y^2/x^2)}{1-(y^2/x^2)}$$ \log x +\log C$
$\log[(x^4 - y^4). (x^2+y^2)^3/x^2-y^2)^3] = 4 \log Cx$
$(x^2+y^2) (x^2+y^2)^3]/(x^2-y^2)^3 = x^4C$
$x^4(x^2+y^2)^4/(x^2-y^2) =C x^4$
$C (x^2 + y^2)^2 = (x^2 - y^2)$
Hence proved.