Solve the following differential equation: 2y e^(x/y) dx + (y - 2x e^(x/y)) dy = 0, given x = 0 and…

$\frac{dx}{dy} = \frac{2x e^{x/y} - y}{2y e^{x/y}}$, which is homogeneous. Put $x = vy$, so $\frac{dx}{dy} = v + y\frac{dv}{dy}$. $v + y\frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} = v - \frac{1}{2e^v}$ $y\frac{dv}{dy} = -\frac{e^{-v}}{2} \Rightarrow 2e^v\,dv = -\frac{dy}{y}$ Integrating: $2e^v = -\log|y| + C$, i.e. $2e^{x/y} + \log|y| = C$. Put $x = 0$, $y = 1$: $2 + 0 = C$, so $C = 2$. Hence $2e^{x/y} + \log y = 2$.