Let y=y(x) be the solution of the differential equation d y= e^α x+y d x; α ∈ N. If y( log_e 2)= log_e 2 and y(0)= log_e (1/2), then the value of α…

Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \mathrm{d} y=\mathrm{e}^{\alpha x+y} \mathrm{~d} x ; \alpha \in \mathrm{N}$. If $\displaystyle y\left(\log _{\mathrm{e}} 2\right)=\log _{\mathrm{e}} 2$ and $\displaystyle y(0)=\log _{\mathrm{e}}\left(\frac{1}{2}\right)$, then the value of $\displaystyle \alpha$ is equal to \_\_\_\_.