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 equation is `(x^3-3xy^2)dx=(y^3-3x^2y)dy`
=> `(dy)/(dx) = (x^3-3xy^2)/(y^3-3x^2y)`
which is a homogeneous equation
therefor put `y=vx`
diff on both side w.r.t x
`(dy)/(dx) = v+ (dv)/(dx)`
=> `v+ (dv)/(dx)= (x^3 - 3xv^2x^2)/(v^3x^3-3x^2.vx)`
`x (dv)/(dx) = (1-3v^2)/(v^3-3v) -v`
` (v^3-3v)/(1-v^4) dv =1/(x) dx`
integrating on both side
`logx = int (v^3-3v)/(1-v^4) dv`
`=> logc +logx = int(v^3)/(1-v^4) dv - 3 int v/(1-v^4) dv`
`log(cx) = I_1 -3I_2`.....(i)
`I_1 = int(v^3)/(1-v^4) dv`
let `1-v^4 = t => dt = -4v^3 dv`
`I_1 = -1/(4) log(1-v^4 )`.....(ii)
`I_2 = int v/(1-v^4) dv`
let `v^2 = t => 2vdv = dt`
`I_2 = int 1/(1-t^2) .((dt)/(2))` = `1/4 log (( 1+t)/(1-t))`
`I_2= 1/4 log (( 1+v^2)/(1-v^2))`
substituting the values of `I_1` and `I_2`
` log(cx) = I_1 = -1/(4) log(1-v^4 ) - 3/4 log (( 1+v^2)/(1-v^2))`
by solving the equation we get
` log(cx) = 1/(4) log[ (1+v^2)^4/ (1-v^2)^2]`
`=> cx = (1-v^2)^(1/2) / (1+v^2)`
`cx = (1-(y/x)^2)^(1/2) / (1+(y/x)^2) => cx= (x^2-y^2)^(1/2) /( x^2+ y^2) `
now
`c( x^2+ y^2)=(x^2-y^2)`
squaring on both side
`c( x^2+ y^2)^2=(x^2-y^2)^2`