Let $$y = y(x)$$ be the solution of the differential equation, $$(x^{2} + 1)^{2}\frac{dy}{dx} + 2x(x^{2} + 1)y = 1$$ such that $$y(0) = 0$$. If $$\sqrt{a} \; y(1) = \frac{\pi}{32}$$, then the value of $$a$$ is:

We have the first-order linear differential equation

$$ (x^{2}+1)^{2}\,\frac{dy}{dx}\;+\;2x\,(x^{2}+1)\,y \;=\;1. $$

The standard form of a linear ODE is $$\displaystyle \frac{dy}{dx}+P(x)\,y=Q(x).$$

To reach that form we divide every term by one power of $$(x^{2}+1)$$ (not two, because that will give the most convenient coefficients):

$$ (x^{2}+1)\,\frac{dy}{dx}\;+\;2x\,y \;=\;\frac{1}{x^{2}+1}. $$

Now we notice that the left-hand side already resembles the derivative of a product. Indeed, the derivative of $$(x^{2}+1)\,y$$ with respect to $$x$$ is

$$\frac{d}{dx}\Bigl[(x^{2}+1)\,y\Bigr] \;=\; (x^{2}+1)\,\frac{dy}{dx} \;+\;2x\,y,$$

which is exactly the left-hand side. Hence the differential equation can be rewritten compactly as

$$ \frac{d}{dx}\Bigl[(x^{2}+1)\,y\Bigr] \;=\; \frac{1}{x^{2}+1}. $$

We now integrate both sides with respect to $$x$$:

$$ \int \frac{d}{dx}\Bigl[(x^{2}+1)\,y\Bigr]\,dx \;=\; \int \frac{dx}{x^{2}+1}. $$

On the left the integral and the derivative cancel, so we obtain

$$ (x^{2}+1)\,y \;=\; \tan^{-1}x \;+\; C, $$

where $$C$$ is the constant of integration.

We are given the initial condition $$y(0)=0$$. Substituting $$x=0$$ and $$y(0)=0$$ into the general solution, we get

$$ (0^{2}+1)\,y(0) \;=\; \tan^{-1}0 \;+\; C \;\;\Longrightarrow\;\; 1\cdot 0 \;=\; 0 \;+\; C \;\;\Longrightarrow\;\; C=0. $$

Thus the particular solution satisfying the initial condition is

$$ y(x)=\frac{\tan^{-1}x}{x^{2}+1}. $$

Now we evaluate this solution at $$x=1$$:

$$ y(1)=\frac{\tan^{-1}1}{1^{2}+1} =\frac{\dfrac{\pi}{4}}{2} =\frac{\pi}{8}. $$

The problem states that

$$ \sqrt{a}\;y(1)=\frac{\pi}{32}. $$

Substituting $$y(1)=\dfrac{\pi}{8}$$ gives

$$ \sqrt{a}\,\Bigl(\frac{\pi}{8}\Bigr) =\frac{\pi}{32}. $$

We cancel $$\pi$$ from both sides:

$$ \frac{\sqrt{a}}{8}=\frac{1}{32}. $$

Multiplying by $$8$$, we obtain

$$ \sqrt{a}=\frac{8}{32} =\frac{1}{4}. $$

Finally, squaring both sides yields

$$ a=\Bigl(\frac{1}{4}\Bigr)^{2} =\frac{1}{16}. $$

Hence, the correct answer is Option A.