Find the particular solution of differential equation
$2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy=0$ given that $x=0\text{ when }y=1.$

Hint: In the given differential equation the equation is in the form of $\dfrac{x}{y}$so, transform the given equation in the form of $\dfrac{dx}{dy}$. After transforming check, the homogeneity of the given equation by putting $x=\lambda x\text{ and }y=\lambda y$. Find the value of $\lambda $, and see its homogeneity. Then solve the differential equation by putting $x=vy$.once we get the standard form, we have to integrate both sides of the differential equation. After integration we get a constant. The value of this constant can be found by substituting $x=0\text{ when }y=1.$

Complete step by step answer:
The given differential equation is
$2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy=0$
This can be written as
$\begin{align}
  & 2y{{e}^{\dfrac{x}{y}}}dx=-\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy \\
 & \Rightarrow 2y{{e}^{\dfrac{x}{y}}}dx=\left( 2x{{e}^{\dfrac{x}{y}}}-y \right)dy \\
 & \Rightarrow \dfrac{dx}{dy}=\dfrac{\left( 2x{{e}^{\dfrac{x}{y}}}-y \right)}{2y{{e}^{\dfrac{x}{y}}}}dy-----(1) \\
\end{align}$
Here we get the differential equation in the form of $\dfrac{dx}{dy}$
Now let us assume that
$F(x,y)=\dfrac{\left( 2x{{e}^{\dfrac{x}{y}}}-y \right)}{2y{{e}^{\dfrac{x}{y}}}}$
Now we have to find $F(\lambda x,\lambda y)$, so we can write
$\begin{align}
  & F(\lambda x,\lambda y)=\dfrac{\left( 2\lambda x{{e}^{\dfrac{\lambda x}{\lambda y}}}-\lambda y \right)}{2\lambda y{{e}^{\dfrac{\lambda x}{\lambda y}}}} \\
 & F(\lambda x,\lambda y)=\dfrac{\lambda \left( 2x{{e}^{\dfrac{\lambda x}{\lambda y}}}-y \right)}{2\lambda y{{e}^{\dfrac{\lambda x}{\lambda y}}}} \\
\end{align}$
Here $\lambda $is cancel out. so, we have
$F(\lambda x,\lambda y)=\lambda {}^\circ F(x,y)$
Hence, we can say that the given differential equation is a homogenous differential equation of zero degree
So, in order to solve this, we have to put
$x=vy$
Hence, we can write
\[\begin{align}
  & \dfrac{dx}{dy}=\dfrac{d}{dy}\left( vy \right) \\
 & \Rightarrow \dfrac{dx}{dy}=\dfrac{ydv}{dy}+\dfrac{vdy}{dy} \\
 & \Rightarrow \dfrac{dx}{dy}=\dfrac{ydv}{dy}+v \\
\end{align}\]
Now we put the value of $\dfrac{dx}{dy}$and $x$in $(1)$we can write further
$\begin{align}
  & \dfrac{dx}{dy}=\dfrac{\left( 2x{{e}^{\dfrac{x}{y}}}-y \right)}{2y{{e}^{\dfrac{x}{y}}}}dy \\
 & \Rightarrow v+y\dfrac{dv}{dy}=\dfrac{2v{{e}^{v}}-1}{2{{e}^{v}}} \\
 & \Rightarrow y\dfrac{dv}{dy}=\dfrac{2v{{e}^{v}}-1}{2{{e}^{v}}}-v \\
 & \Rightarrow y\dfrac{dv}{dy}=\dfrac{2v{{e}^{v}}-1-2v{{e}^{v}}}{2{{e}^{v}}} \\
 & \Rightarrow y\dfrac{dv}{dy}=\dfrac{-1}{2{{e}^{v}}} \\
\end{align}$
Now on cross multiplication we can write
$2{{e}^{v}}dv=\dfrac{-dy}{y}$
now at this step we can integrate easily, so, integrating both side we can write
$\begin{align}
  & \int{2{{e}^{v}}}dv=\int{\dfrac{-dy}{y}} \\
 & \Rightarrow 2\int{{{e}^{v}}}dv=-\int{\dfrac{dy}{y}} \\
 & \Rightarrow 2{{e}^{v}}=-\log \left| y \right|+c \\
\end{align}$
Here we use the formula$\int{{{e}^{x}}}dx={{e}^{x}}+{{c}_{1}}\text{ and }\int{\dfrac{1}{y}}dy=\log y+{{c}_{2}}$
Here $c$is the combined constant of ${{c}_{1}}\text{ and }{{c}_{2}}$
Now we put the value of $v=\dfrac{x}{y}$, hence we can write
$2{{e}^{\dfrac{x}{y}}}+\log \left| y \right|=c-----(2)$
Now from question we have the value $x=0$when $y=1$, so we have to put these values in order to find the value of $c$, hence we can write
$\begin{align}
  & 2{{e}^{\dfrac{0}{1}}}+\log \left| 1 \right|=c \\
 & \Rightarrow c=2 \\
\end{align}$
As we know that ${{e}^{0}}=1\text{ and }\log 1=0$
Now we put the value of $c\text{ in }(2)$, we can write
$2{{e}^{\dfrac{x}{y}}}+\log \left| y \right|=2$
So, this is the particular solution.

Note:
A function $f(x,y)$in $x\text{ and }y$is said to be a homogeneous function of degree $n$, if the degree of each term is $n$. For solving the homogeneous equation, we put $y=vx$and differentiate it and proceed.