### Differential Equations Question 4

#### Question 4 - 25 January - Shift 2

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 $Lim _{t \to \infty} y(t)$

(1) is 0

(2) does not exist

(3) is 1

(4) is -1

## Show Answer

#### Answer: (1)

#### Solution:

#### Formula: Linear differential equations of first order

$\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t}$

I.F. $=e^{\int \alpha dt}=e^{\alpha t}$

Solution $\Rightarrow y \cdot e^{\alpha t}=\int \gamma e^{-\beta t} \cdot e^{\alpha t} d t$

$\Rightarrow ye^{\alpha t}=\gamma \frac{e^{(\alpha-\beta) t}}{(\alpha-\beta)}+c$

$\Rightarrow y=\frac{\gamma}{e^{\beta t}(\alpha-\beta)}+\frac{c}{e^{\alpha t}}$

So, $\lim _{t \to \infty} y(t)=\frac{\gamma}{\infty}+\frac{c}{\infty}=0$