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(B) Given the differential equation: $dy = e^{\alpha x + y} dx$.Rearranging the terms to separate the variables:$e^{-y} dy = e^{\alpha x} dx$.Integrat

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$2$

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(B) Given the differential equation: $dy = e^{\alpha x + y} dx$.Rearranging the terms to separate the variables:$e^{-y} dy = e^{\alpha x} dx$.Integrating both sides:$\int e^{-y} dy = \int e^{\alpha x} dx$$-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C \quad \dots(i)$Using the condition $y(0) = \log_{e}(\frac{1}{2}) = -\log_{e} 2$:$-e^{-(-\log_{e} 2)} = \frac{e^{\alpha(0)}}{\alpha} + C$$-e^{\log_{e} 2} = \frac{1}{\alpha} + C$$-2 = \frac{1}{\alpha} + C \quad \dots(ii)$Using the condition $y(\log_{e} 2) = \log_{e} 2$:$-e^{-\log_{e} 2} = \frac{e^{\alpha \log_{e} 2}}{\alpha} + C$$-\frac{1}{2} = \frac{2^{\alpha}}{\alpha} + C \quad \dots(iii)$Subtracting equation $(ii)$ from equation $(iii)$:$(-\frac{1}{2}) - (-2) = \frac{2^{\alpha}}{\alpha} - \frac{1}{\alpha}$$\frac{3}{2} = \frac{2^{\alpha} - 1}{\alpha}$For $\alpha = 2$:$\frac{2^{2} - 1}{2} = \frac{4 - 1}{2} = \frac{3}{2}$.Thus,the value of $\alpha$ is $2$.

Let $y=y(x)$ be the solution of the differential equation $dy=e^{\alpha x+y} dx$; $\alpha \in N$. If $y(\log_{e} 2)=\log_{e} 2$ and $y(0)=\log_{e}(\frac{1}{2})$,then the value of $\alpha$ is equal to $.....$

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