Question

For $\alpha \in ℝ$ , let $y_{\alpha }\left(x\right)$ be the solution of the differential equation

$\frac{dy}{dx}+2y=\frac{1}{1+x^2}\text{for}x\in ℝ$

satisfying $y\left(0\right)=\alpha$ . Then, which one of the following is TRUE?

1. $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)=0$ for every $\alpha \in ℝ$
2. $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)=1$ for every $\alpha \in ℝ$
3. There exists an $\alpha \in ℝ$ such that $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)$ exists but its value is different from 0 and 1
4. There is an $\alpha \in ℝ$ for which $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)$ does not exist

Answer: $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)=0$ for every $\alpha \in ℝ$

Solution :-

We have differential equation $\frac{dy}{dx}+2y=\frac{1}{1+x^2}$ this is a first order first degree linear differential equation.
Here, $P\left(x\right)=2$ and $Q\left(x\right)=\frac{1}{1+x^2}\mathrm{.}$

So, $\text{I.F.}=e^{\int P\left(x\right)dx}=e^{\int 2dx}=e^{2x}\mathrm{.}$

Now, the solution of the given differential equation is

Let $e^{2x}=t$ and taking log both sides, we get $2x=logt⟹x=\frac{logt}{2}\mathrm{.}$
Now, differentiate $2x=logt$ w. r. to $x$ , we get

Now, from equation (1) we have

Now, given that $y\left(0\right)=\alpha$ so from eq(2), we have

From eq(2), we get

Here, ${tan}^{-1}x$ goes to $\frac{\pi }{2}$ and $x\to \infty$ and $e^{2x}$ goes to larger and larger as $x\to \infty$ so $\underset{x\to \infty }{lim}\frac{{tan}^{-1}x}{e^{2x}}=0.$

Also, $\underset{x\to \infty }{lim}\frac{\alpha }{e^{2x}}$ goes to 0 as $x\to \infty$ because $\alpha$ is any fixed real number.

So, for every $\alpha \in ℝ,\underset{x\to \infty }{lim}\frac{\alpha }{e^{2x}}=0.$

Therefore, from eq(3) we get $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)=0+0=0.$

So, the correct answer is option-1. $\to$ $\underset{x\to \infty }{lim}{y}_{\alpha }\left(x\right)=0$ for every $\alpha \in ℝ$