Question

Let $y=y(t)$ be a solution of the differential equation $\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$ where, $\alpha >0,\beta >0$ and $\gamma >0$. Then $\underset{t\to \infty }{\lim }y(t)$

• A. is$-1$
• B. is 0
• C. is 1
• D. does not exist

Answer 1

Answer: (B)

$\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$
$\text{ I.F. }=e^{\int \alpha dt}=e^{\alpha t}$
Solution $\Rightarrow y\cdot e^{\alpha t}=\int \gamma c^{-\beta T}\cdot c^{\alpha t}dt$
$\Rightarrow y{e}^{\alpha t}=\gamma \frac{e^{(\alpha -\beta )t}}{(\alpha -\beta )}+c$
$\Rightarrow y=\frac{\gamma }{e^{\beta t}(\alpha -\beta )}+\frac{c}{e^{\alpha t}}$
So, $\underset{t\to \infty }{\lim }y(t)=\frac{\gamma }{\infty }+\frac{c}{\infty }=0$