Mathematics - Differential Equation Question with Solution | TestHub

We have, $\int e^{-y}dy=\int e^{\alpha x}dx$

$\Rightarrow -e^{-y}=\frac{e^{\alpha x}}{\alpha }+c ...\left(1\right)$ 

Put $\left(x,y\right)=\left(\ell n2,\ell n2\right)$

$-\frac{1}{2}=\frac{2^{\alpha }}{\alpha }+c ...\left(2\right)$

Put $\left(x,y\right)\equiv \left(0,-\ell n2\right)$ 

$-2=\frac{1}{\alpha }+c$

From equation $\left(1\right)-\left(2\right),$ we get

$\frac{2^{\alpha }-1}{\alpha }=\frac{3}{2}$

$\Rightarrow \alpha =2($ as $\alpha \in N)$.