Exams Joint Entrance Examination Mathematics Differential Equations Question 453 Question 453 of 490 Medium MCQ
Let y = y ( t ) y=y(t) 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 \[\mathop {\lim }\limits_{t \to \infty } y(t)\]
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d y d t + α y = γ e − β t \frac{d y}{d t}+\alpha y=\gamma e^{-\beta t} d t d y + α y = γ e − β t
\[
I.F. = e ∫ α d t = e α t ⇒ y ⋅ e α t = γ ∫ e ( α − β ) t d t = γ e ( α − β ) t ( α − β ) + C ⇒ y = γ ( α − β ) e − β t + C e − α t lim x → ∞ y ( t ) = lim x → ∞ [ γ ( α − β ) e − β t + C e − α t ] = 0 \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} I.F. = e ∫ α d t = e α t ⇒ y ⋅ e α t = γ ∫ e ( α − β ) t d t = γ ( α − β ) e ( α − β ) t + C ⇒ y = ( α − β ) γ e − β t + C e − α t x → ∞ lim y ( t ) = x → ∞ lim [ ( α − β ) γ e − β t + C e − α t ] = 0
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