Question

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

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

Answer 1

Answer: (B)

$ \frac{d y}{d t}+\alpha y=\gamma e^{-\beta t} $
$ \text { I.F. }=e^{\int \alpha d t}=e^{\alpha t} $
Solution $ \Rightarrow y \cdot e^{\alpha t}=\int \gamma c^{-\beta T } \cdot c^{\alpha t} d t $
$ \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, $\displaystyle\lim _{t \rightarrow \infty} y(t)=\frac{\gamma}{\infty}+\frac{c}{\infty}=0$