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