Let $y=y(t)$ be a solution of the differential equation ${{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}}$ where, $\alpha > 0,\beta > 0$ and $\gamma > 0$. Then $\mathop {\lim }\limits_{t \to \infty } y(t)$
Solution
$\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$
<br/><br/>
$$
\begin{aligned}
& \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\
& \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\
& \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\
& \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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