Let y=y(t) be a solution of the differential equation dy dt+α y=γ e^-β t Where, α >0,β >0 and γ >0. Then Lim_t→ ∞ y(t)

Given,

$$\frac{dy}{dt}+\alpha y=\gamma \cdot {e}^{-\beta t}$$dtdy​+αy=γ⋅e−βt

Which is a linear differential equation,

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

So, the solution of differential equation is given by

$$y\times {e}^{\alpha t}=\gamma \int {e}^{-\beta t}\times {e}^{\alpha t}$$y×eαt=γ∫e−βt×eαt

$$\Rightarrow y\times {e}^{\alpha t}=\gamma \int {e}^{(\alpha -\beta )t}$$⇒y×eαt=γ∫e(α−β)t

$$\Rightarrow y\times {e}^{\alpha t}=\gamma \frac{{e}^{(\alpha -\beta )t}}{(\alpha -\beta )}+c$$⇒y×eαt=γ(α−β)e(α−β)t​+c

$$\Rightarrow y=\gamma \frac{{e}^{-\beta t}}{(\alpha -\beta )}+c{e}^{-\alpha t}$$⇒y=γ(α−β)e−βt​+ce−αt

Now finding $\underset{t\rightarrow \infty }{\mathrm{lim}}y$t→∞lim​y we get,

$$\underset{t\rightarrow \infty }{\mathrm{lim}}y=\underset{t\rightarrow \infty }{\mathrm{lim}}(\gamma \frac{{e}^{-\beta t}}{(\alpha -\beta )}+c{e}^{-\alpha t})$$t→∞lim​y=t→∞lim​(γ(α−β)e−βt​+ce−αt)

$$\Rightarrow \underset{t\rightarrow \infty }{\mathrm{lim}}y=\gamma \times 0+c\times 0=0{\text{as }{e}^{-\infty }\rightarrow 0}$$⇒t→∞lim​y=γ×0+c×0=0as e−∞→0