Particular solution of $2ye^{\tfrac{x}{y}}dx+\Big(y-2xe^{\tfrac{x}{y}}\Big)dy=0$, $x=0$ when $y=1$
Attempt
Put $x=vy$ $$ \frac{dx}{dy}=\frac{x}{y}-\frac{1}{2e^{\tfrac{x}{y}}}\\ \frac{dx}{dy}=v+y\frac{dv}{dy}=v-\frac{1}{2e^{v}}\implies y\frac{dv}{dy}=-\frac{1}{2e^{v}}\\ \implies\int2e^{v}dv=-\int\frac{dy}{y}\implies2e^{v}=-\log|y|+C\\ 2e^{v}+\log|y|=C\implies\color{red}{2e^{\tfrac{x}{y}}+\log|y|=C} $$ $x=0$ when $y=1\implies C=2$ $$ {2e^{\tfrac{x}{y}}+\log|y|=2} $$ which is what is given in my reference as solution, but can the modulus function be involved in the particular solution of a differential equation ?
Or do I need to eliminate the solution $2e^{\tfrac{x}{y}}+\log({\text{-}y})=2$ ?
$$ 2e^{\tfrac{x}{y}}+\log|y|=C\implies\begin{cases}2e^{\tfrac{x}{y}}+\log y=C\\ 2e^{\tfrac{x}{y}}+\log(\text{-}y)=C\end{cases} $$ The condition "$x=0$ when $y=1$" is only applicable to case 1 as $y=1>0$. Thus, $$ 2e^{\frac{0}{1}}+\log 1=C\implies C=2\\ 2e^{\tfrac{x}{y}}+\log y=2 $$ is the particular solution when $x=0;y=1$.