JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 14)
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 $$\mathop {\lim }\limits_{t \to \infty } y(t)$$
is 0
is 1
is $$-1$$
does not exist
Explanation
$\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}$
$$ \begin{aligned} & \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\ & \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\ & \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\ & \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0 \end{aligned} $$
$$ \begin{aligned} & \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} \\\\ & \Rightarrow y \cdot e^{\alpha t}=\gamma \int e^{(\alpha-\beta) t} d t=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+C \\\\ & \Rightarrow y=\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t} \\\\ & \lim _{x \rightarrow \infty} y(t)=\lim _{x \rightarrow \infty}\left[\frac{\gamma}{(\alpha-\beta)} e^{-\beta t}+C e^{-\alpha t}\right]=0 \end{aligned} $$
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