Let y = y ( t ) be a solution of the differential equation d y d t + y = e - t where, 0 , 0 and 0 . Then lim t y ( t ) [2023]

Question

Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

Smart Achievers · Accepted Answer

(2)

$\frac{d y}{d t} + α y = γ e^{- β t}$

$I.F. = e^{\int α d t} = e^{α t}$

Solution is given by

$y \cdot e^{α t} = \int e^{α t} \cdot γ e^{- β t} d t = γ \int e^{(α - β) t} d t$

$\Rightarrow y \cdot e^{α t} = γ \frac{e^{(α - β) t}}{α - β} + c$ $\Rightarrow y = \frac{γ}{α - β} e^{- β t} + c e^{- α t}$

$So, \lim_{t \to \infty} y (t) = \frac{γ}{α - β} \lim_{t \to \infty} e^{- β t} + c \lim_{t \to \infty} e^{- α t} = 0$

Solution

Q.
Let $y = y (t)$ be a solution of the differential equation $\frac{d y}{d t} + α y = γ e^{- β t}$ where, $α > 0, β > 0$ and $γ > 0$ . Then $\lim_{t \to \infty} y (t)$ [2023]

1 is -1

2 is 0

3 is 1

4 does not exist

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