Answer
\[y(x)=\frac{e^{\beta x}}{\alpha+\beta}+Ce^{-\alpha x}\]
Work Step by Step
$y'+\alpha y=e^{\beta x}$ _____(1)
Integrating factor
\[I(x)=e^{\int\alpha dx}=e^{\alpha x}\]
Multiplying (1) by $I(x)$
\[e^{\alpha x}\frac{dy}{dx}+\alpha e^{\alpha x}y=e^{(\alpha+\beta) x}\]
\[\frac{d}{dx}(ye^{\alpha x})=e^{(\alpha+\beta) x}\]
Integrating,
\[ye^{\alpha x}=\int e^{(\alpha+\beta) x}dx+C\]
Where $C$ is constant of integration
\[ye^{\alpha x}=\frac{e^{(\alpha+\beta) x}}{\alpha+\beta}+C\]
\[y(x)=\frac{e^{\beta x}}{\alpha+\beta}+Ce^{-\alpha x}\]
Hence general solution is $y(x)=\Large\frac{e^{\beta x}}{\alpha+\beta}+\large Ce^{-\alpha x}$
