The general solution of the differential equation y2+e2xdy−y3dx=0, (c being the constant  of integration), is

We have dxdy=y2+e2xy3⇒dxdy=1y+e2xy3⇒e−2xdxdy−e−2x1y=1y3→1Let −e−2x2=u→2⇒e2xdxdy=dudy→3Substituting 2,3 in 1, we getdudy+2yu=1y3whose solution is u.y2=∫1ydy+k⇒uy2=ln⁡y+k⇒−e−2x2y2=ln⁡y+k   ∵u=−e−2x2⇒−e−2xy2=2ln⁡y+2k∴e−2xy2+2ln⁡y=c where c is an arbitrary constant