Using homogeneous method
let y=vx⇒dxdy=x2y(x3+y3)dxdy=x2(vx)(x3+(vx)3)⇒dxdy=x3v(x3+v3x3)dxdy=x3vx3(1+v3)dxdy=v(1+v3)Therefordxdy=dx1+vdv⇒dx1+vdv=v(1+v3)⇒dxvdv=v(−1)(1+v3)dxdy=(v+v21)−1dxdy=(1/v2+v−1/v)⇒(v2+v∧−1−v)dv=dx⇒(3v3)+(lnv)−(2v2)=xbutv=xy⇒3(xy)3+(ln(xy))−2(xy)2=x+c
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