For a differential function $\varphi \text{ with }\varphi \left(1\right)=2$ the solution of the differential equation$y^{\mathrm{\prime }}=\frac{y}{x}+\frac{\varphi (y/x)}{{\varphi }^{\mathrm{\prime }}(y/x)},y\left(2\right)=2$ is.

Putting $y/x=u,\text{ we have }\frac{\mathrm{d}y}{\mathrm{d}x}=u+x\frac{\mathrm{d}u}{\mathrm{d}x}$ then given differential equation can be written as

$$\begin{gathered} u+x\frac{\mathrm{d}u}{\mathrm{d}x}=u+\frac{\varphi \left(u\right)}{{\varphi }^{\mathrm{\prime }}\left(u\right)}\Rightarrow x\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\varphi \left(u\right)}{{\varphi }^{\mathrm{\prime }}\left(u\right)} \\ \Rightarrow \frac{{\varphi }^{\mathrm{\prime }}\left(u\right)}{\varphi \left(u\right)}\mathrm{d}u=\frac{\mathrm{d}x}{x} \end{gathered}$$

integrating we get  $\log \varphi \left(u\right)=\log x+\log k,\text{ so }\varphi \left(u\right)=kx\text{ i.e., }\varphi (y/x)=kx.\text{ Putting }x=2,y=2,k=\frac{1}{2}\varphi .\left(1\right)=1\text{. }$