Let $y=y(x)$ be the solution of the differential equation $d y=e^{\alpha x+y} d x ; \alpha \in \mathbf{N}$. If $y\left(\log _{e} 2\right)=\log _{e} 2$ and $y(0)=\log _{e}\left(\frac{1}{2}\right)$, then the value of $\alpha$ is equal
to______________
$\int e^{-y} d y=\int e^{\alpha x} d x$
$\Rightarrow \mathrm{e}^{-\mathrm{y}}=\frac{\mathrm{e}^{\alpha \mathrm{x}}}{\alpha}+\mathrm{c}$ .......(I)
Put $(\mathrm{x}, \mathrm{y})=(\ln 2, \ell \mathrm{n} 2)$
$\frac{-1}{2}=\frac{2^{\alpha}}{\alpha}+C$ ..............(II)
Put $(\mathrm{x}, \mathrm{y}) \equiv(0,-\ell \mathrm{n} 2)$ in (I)
$-2=\frac{1}{\alpha}+C$ ...................(III)
(ii) - (iii)
$\frac{2^{\alpha}-1}{\alpha}=\frac{3}{2}$
$\Rightarrow \alpha=2($ as $\alpha \in \mathbb{N})$
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