Let us solve the equation dxdy=−xyx2+y2 using the transformation y=ux. Then dxdy=dxdux+u, and hence we get the equation dxdux+u=−u1+u2, which is equivalent to dxdux=−u1+u2−u, and hence to dxdux=−u1+2u2.
It follows that
1+2u2udu=−xdx
∫1+2u2udu=−∫xdx
41∫1+2u24udu=−∫xdx
∫1+2u2d(1+2u2)=−4∫xdx
ln(1+2u2)=−4ln∣x∣+ln∣C∣
ln(1+2u2)+4ln∣x∣=ln∣C∣
ln(x4(1+2u2))=ln∣C∣
x4(1+2u2)=C
x4(1+2(xy)2)=C
We conclude that the general solution of the equation dxdy=−xyx2+y2 is
x4+2x2y2=C
Comments