x[x2+y2]dydx=y[y2−x2], ⇒ dydx=y3−yx2x3+y2.......[1]
Dividing numerator and denominator on the right hand side of .....[1] by x3 and substituting v=yx will give a homogeneous equation in v of the form.......
dydx=v3−v1+v2 . From v=yx, ⇒ y=vx.......[2]
Differentiating.....[2] w.r.t x [implicitly], dydx=xdvdx+v and sustituting for dydx we obtain,
xdvdx = v3−v1+v2−v, i.e, xdvdx=−2v1+v2 which will yield
dxx=−1+v22vdv , therefore, −dxx=[12v+v22v]dv.
Integrating both sides, −lnx=12lnv+v24+C and multiplying both sides by 2 and let C =lnA will give,
−lnx2=lnv+v22+lnA, tidying up, ln[Ax2v]=−v22,
Ax2v = e−[v22][ where 1A is equal to another constant B given in the answer ] and so,
Substituting back for v=yx, we will obtain the answer given above. I hope this was both helpful and correct, and will ask for it to be checked.
