Show that the differential equation $2 y e^{\frac{x}{y}} dx + (y - 2 x e^{\frac{x}{y}}) dy = 0$ is homogeneous and find its particular solution,given that $x = 0$ when $y = 1$.

(C) The given differential equation can be written as:
$\frac{dx}{dy} = \frac{2 x e^{\frac{x}{y}} - y}{2 y e^{\frac{x}{y}}}$ ........... $(1)$
Let $F(x, y) = \frac{2 x e^{\frac{x}{y}} - y}{2 y e^{\frac{x}{y}}}$.
Then $F(\lambda x, \lambda y) = \frac{\lambda(2 x e^{\frac{x}{y}} - y)}{\lambda(2 y e^{\frac{x}{y}})} = \lambda^0 F(x, y)$.
Thus,$F(x, y)$ is a homogeneous function of degree zero. Therefore,the given differential equation is a homogeneous differential equation.
To solve it,we make the substitution $x = vy$ ........... $(2)$.
Differentiating equation $(2)$ with respect to $y$,we get $\frac{dx}{dy} = v + y \frac{dv}{dy}$.
Substituting the value of $x$ and $\frac{dx}{dy}$ in equation $(1)$,we get:
$v + y \frac{dv}{dy} = \frac{2 v e^v - 1}{2 e^v}$
$y \frac{dv}{dy} = \frac{2 v e^v - 1}{2 e^v} - v$
$y \frac{dv}{dy} = -\frac{1}{2 e^v}$
$2 e^v dv = -\frac{dy}{y}$
Integrating both sides,we get $\int 2 e^v dv = -\int \frac{dy}{y}$.
$2 e^v = -\log |y| + C$.
Replacing $v$ by $\frac{x}{y}$,we get $2 e^{\frac{x}{y}} + \log |y| = C$ ........... $(3)$.
Substituting $x = 0$ and $y = 1$ in equation $(3)$,we get $2 e^0 + \log |1| = C \Rightarrow C = 2$.
Substituting the value of $C$ in equation $(3)$,the particular solution is $2 e^{\frac{x}{y}} + \log |y| = 2$.