.......(i) This is a homogeneous equation ∴ we put y=vx and
dy
dx
=v+x
dv
dx
The Eq. (i) is reduces to v+x
dv
dx
=
x2(1+v2)
2x2v
⇒ x
dv
dx
=
1+v2
2v
−v=
1−v2
2v
⇒ −
2v
1−v2
dv=−
dx
x
On integrating both sides, we get log(1−v2)=−log‌x+log‌c ⇒log(x2−y2)−2‌log‌x=−logx+log‌c ⇒log(x2−y2)=log‌x‌c ⇒x2−y2=xc