(x2−y2)dx=2xydy
or, dydx=x2−y22xy ............ (1)
Let y=vx⇒dydx=v+xdvdx
∴ From (1),v+xdvdx=1−v22v
or, xdvdx=1−v22v−v=1−3v22v
or, v1−3v2dv=dxx
Integrating both sides,
∫vdv1−3v2=∫dxx+c ......... (2) c= constant of integrating.
Let 1−3v2=z⇒−6vdv=d⇒vdv=−16dz
∴ (2)
∫−16dzz=∫dxx+c
or, −16logz=logx+c
or, −log(1−3v2)=6logx+6c
or, logx6+log(1−3v2)=−6c
or, logx6(1−3v2)=−6c
or, x6(1−3v2)=e−6c=k k= constant of integration
or, x6(1−3y2x2)=k
∴x4(x2−3y2)=4⇒ Required general solution.