Question

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

Answer 1

$\frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}$

Let $\frac{y}{x}=v\Rightarrow y=vx$

$y^1=v+x{v}^1$

$v+x{v}^1=\frac{1-3v^2}{v^3-3v}$

$x{v}^1=\frac{1-3v^2-v}{v^3-3v}\Rightarrow x{v}^1=\frac{1-v^4}{v^3-3v}$

$=\int \frac{v^3-3v}{1-v^4}dv=\int \frac{1}{x}dx$

$\frac{-1}{4}\int \frac{v^3}{1-v^4}dv-\int \frac{3v}{1-v^4}dv=\int \frac{1}{x}dx$

$\left[\frac{-1}{4}log\right(1-v^4)-\frac{1}{4}log(\frac{1+v^2}{1-v^2}\left)\right]=logx+logC$

$\frac{-1}{4}log\left[\frac{(1+v^2{)}^4}{(1-v^2{)}^2}\right]=logCx$

$log[\frac{(1-v^2{)}^2}{(1+v^2{)}^4}{]}^{1/4}=logCx$

$Cx=[\frac{(1-v^2{)}^2}{(1+v^2{)}^4}{]}^{1/4}=\frac{(1-v^2{)}^{1/2}}{(1+v^2)}$

$xC(1+v^2)=(1-v^2{)}^{1/2}$

$C\left(\frac{x^2+y^2}{x^2}\right)x=(\frac{x^2-y^2}{x^2}{)}^{1/2}$

$C\left(\frac{x^2+y^2}{x}\right)=\frac{\sqrt{x^2-y^2}}{x}$

squaring both sides, we get,

$(x^2-y^2)=C(x^2+y^2{)}^2$

Which is the required solution.