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.

Explanation

Given, `(x^(3)3xy^(2))dx=(y^(3)-3x^(2)y)dy`
`(dy)/(dx)=(x^(3)-3xy^(2))/(y^(3)-3x^(2)y)`
`implies v+x(dv)/(dx)=(x^(3)-3x^(3)v^(2))/(v^(3)x^(3)-3x^(3)v)`
Let `y=vx`
`implies (dy)/(dx)=v+x(dv)/(dx)`
`=(1-3v^(2))/(v^(3)-3v)`
`implies x(dv)/(dx)=(1-3v^(2))/(v^(3)-3v)-v`
`=(1-3v^(2)-v^(4)+3v^(2))/(v^(3)-3v)`
`=(1-v^(4))/(v^(3)-3v)`
`implies (v^(3)-3v)/(1-v^(4))dv=(dx)/(x)`
`implies int(2v(v(2)-3))/(1-v^(4))dv=2int(dx)/(x)+logc`
`implies int(t-3)/(1-t^(2))dt=2logx+logc` (Let `v^(2)=timplies2vdv=dt`)
`=int{(-1)/(1-t)-(2)/(1+t)}dt=2logx+logc` (Using partial fractions)
`implies (log(1-t)-2log(1+t)=log(cx^(2))`
`implieslog(1-t)/((1+t)^(2))=logcx^(2))`
`implies (1-t)/((1+t)^(2))=cx^(2)`
`implies1-v^(2)=cx^(2)(1+v^(2))^(2)`
`implies 1-(y^(2))/(x^(2))=cx^(2)(1+(y^(2))/(x^(2)))^(2)`
`implies x^(2)-y^(2)=c(x^(2)+y^(2))^(2)`