(x^3 +y^3)dx-3xy^2dy=0

Question

find the general soulution/ particular solution

Answer 1

Let y=vx, then dy/dx=v+xdv/dx. The DE becomes:

x³+v³x³-3v²x³(v+xdv/dx)=0,

1+v³-3v³=3v²xdv/dx,

xdv/dx=(1-2v³)/(3v²),

(3v²/(1-2v³))dv=dx/x, integrating:

∫(3v²/(1-2v³))dv=ln(Ax) where A is a constant.

Let u=1-2v³, then du/dv=-6v², dv=-du/6v²,

∫(3v²/u)(-du/6v²)=ln(Ax),

-½∫du/u=-½ln(u)=ln(Ax),

ln(u)=-2ln(Ax)=ln(C/x²), where C is a constant,

u=C/x²,

1-2v³=C/x²,

1-2y³/x³=C/x²,

x³-2y³=Cx, x³-2y³-Cx=0.