Given,

$\frac{dy}{dt}+\alpha y=\gamma ·e^{-\beta t}$

Which is a linear differential equation,

So, integrating factor will be $I.F=e^{\int \alpha dt}=e^{\alpha t}$

So, the solution of differential equation is given by

$y\times e^{\alpha t}=\gamma \int e^{-\beta t}\times e^{\alpha t}$

$\Rightarrow y\times e^{\alpha t}=\gamma \int e^{\left(\alpha -\beta \right)t}$

$\Rightarrow y\times e^{\alpha t}=\gamma \frac{e^{\left(\alpha -\beta \right)t}}{\left(\alpha -\beta \right)}+c$

$\Rightarrow y=\gamma \frac{e^{-\beta t}}{\left(\alpha -\beta \right)}+c{e}^{-\alpha t}$

Now finding $\underset{t\to \infty }{\lim }y$ we get,

$\underset{t\to \infty }{\lim }y=\underset{t\to \infty }{\lim }\left(\gamma \frac{e^{-\beta t}}{\left(\alpha -\beta \right)}+c{e}^{-\alpha t}\right)$

$\Rightarrow \underset{t\to \infty }{\lim }y=\gamma \times 0+c\times 0=0 \left\{\text{as }{e}^{-\infty }\to 0\right\}$