$2\mathit{y}{\mathit{e}}^{\mathit{x}∕\mathit{y}}$ dx + (y - 2x ${\mathit{e}}^{\mathit{x}∕\mathit{y}}$) dy = 0
$\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{y}}$ = $\frac{2\mathit{x}{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}-\mathit{y}}{2\mathit{y}{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}}$ ... (1)
Let f (x , y) = $\frac{2\mathit{x}{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}-\mathit{y}}{2\mathit{y}{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}}$
Then, (λx , λy) = $\frac{\mathit{\lambda }\left(2\mathit{x}{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}-\mathit{y}\right)}{\mathit{\lambda }\left(2\mathit{y}{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}\right)}$ = ${\mathit{\lambda }}^0$ [F (x,y)]
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
Let x = vy
Differentiating w.r.t. y, we get
$\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{y}}$ = v + y $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{y}}$
Substituting the value of x and $\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{y}}$ in equation (1), we get
v + y $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{y}}$ = $\frac{2\mathit{v}\mathit{y}{\mathit{e}}^{\mathit{v}}-\mathit{y}}{2\mathit{y}{\mathit{e}}^{\mathit{v}}}$ = $\frac{2\mathit{v}{\mathit{e}}^{\mathit{v}}-1}{2{\mathit{e}}^{\mathit{v}}}$
or y $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{y}}$ = $\frac{2\mathit{v}{\mathit{e}}^{\mathit{v}}-1}{2{\mathit{e}}^{\mathit{v}}}$ - v
or y $\frac{\mathit{d}\mathit{v}}{\mathit{d}\mathit{y}}$ = - $\frac{1}{2{\mathit{e}}^{\mathit{v}}}$
or $2{\mathit{e}}^{\mathit{v}}$ dv = - $\frac{\mathit{d}\mathit{y}}{\mathit{y}}$
or ∫ $2{\mathit{e}}^{\mathit{v}}$ . dv = - ∫ $\frac{\mathit{d}\mathit{y}}{\mathit{y}}$
or $2{\mathit{e}}^{\mathit{v}}$ = - log |y| + C
Substituting the value of v, we get
$2{\mathit{e}}^{\frac{\mathit{\pi }}{\mathit{y}}}$ + log |y| = C ... (2)
Substituting x = 0 and y = 1 in equation (2), we get
$2{\mathit{e}}^0$ + log |1| = C ⇒ C = 2
Substituting the value of C in equation (2), we get
$2{\mathit{e}}^{\frac{\mathit{x}}{\mathit{y}}}$ + log |y| = 2, which is the particular solution of the given differential equation.