If the general solution of the differential equation y^prime= y x+Φ ( x y ), for some function Φ, is given by y ln |c x|=x, where is an arbitrary…

Given $\frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{y}{x}\right)$dxdy​=xy​+ϕ(xy​) Let $\left(\frac{y}{x}\right)=v$(xy​)=v so that $y=x v$y=xv or $\frac{d y}{d x}=x \frac{d v}{d x}+v$dxdy​=xdxdv​+v from (1) & (2), $x \frac{d v}{d x}+v=v+\phi\left(\frac{1}{v}\right)$xdxdv​+v=v+ϕ(v1​) or, $\frac{d v}{\phi\left(\frac{1}{v}\right)}=\frac{d x}{x}$ϕ(v1​)dv​=xdx​ Integrating both sides, we get $\displaystyle \int \frac{d x}{x}=\int \frac{d v}{\phi\left(\frac{1}{v}\right)} \Rightarrow \ln x+c=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$∫xdx​=∫ϕ(v1​)dv​⇒lnx+c=∫ϕ(v1​)dv​ (where $\mathrm{c}$c being constant of integration) But, given $\mathrm{y}=\frac{x}{\ln |c x|}$y=ln∣cx∣x​ is the general solution so that $\displaystyle \frac{x}{y}=\frac{1}{v}=\ln |c x|=\int \frac{d v}{\phi\left(\frac{1}{v}\right)}$yx​=v1​=ln∣cx∣=∫ϕ(v1​)dv​ Differentiating w.r.t $v$v both sides, we get $\phi\left(\frac{1}{v}\right)=\frac{-1}{v^2} \Rightarrow \phi\left(\frac{x}{y}\right)=-\frac{y^2}{x^2}$ϕ(v1​)=v2−1​⇒ϕ(yx​)=−x2y2​ when $\frac{x}{y}=2$yx​=2 i.e. $\phi(2)$ϕ(2) $$=-\left(\frac{y}{x}\right)^2=-\left(\frac{1}{2}\right)^2=\left(\frac{-1}{4}\right)$$=−(xy​)2=−(21​)2=(4−1​)