Given dxdy=xy+ϕ(xy) Let (xy)=v so that y=xv or dxdy=xdxdv+v from (1) & (2), xdxdv+v=v+ϕ(v1) or, ϕ(v1)dv=xdx Integrating both sides, we get ∫xdx=∫ϕ(v1)dv⇒lnx+c=∫ϕ(v1)dv (where c being constant of integration) But, given y=ln∣cx∣x is the general solution so that yx=v1=ln∣cx∣=∫ϕ(v1)dv Differentiating w.r.t v both sides, we get ϕ(v1)=v2−1⇒ϕ(yx)=−x2y2 when yx=2 i.e. ϕ(2) =−(xy)2=−(21)2=(4−1)