The solution of $x\frac{dy}{dx}=y+x{e}^{y/x}\text{ with }y\left(1\right)=0$ is.

Put $y=vx.\text{ Then }\frac{dy}{dx}=v+x\frac{dv}{dx}$

Given equation is $x\frac{dy}{dx}=y+x{e}^{y/x}\Rightarrow \frac{dy}{dx}=\frac{y}{x}+e^{y/x}\Rightarrow v+x\frac{dv}{dx}=v+e^v$

$\begin{array}{l}\Rightarrow x\frac{dv}{dx}=e^v\Rightarrow e^{-v}dv=\frac{dx}{x}\Rightarrow \int e^{-v}dv=\int \frac{dx}{x}\Rightarrow -e^{-v}=\log x-c\Rightarrow \log x+e^{-v}=c\\ \Rightarrow \log x+e^{y/x}=c\Rightarrow y\left(1\right)=0\Rightarrow \log 1+e^0=c\Rightarrow c=1\end{array}$

$\therefore$The required solution is $e^{y/x}+\log x=1$