JEE Math Practice Question

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

1. A.$${1 \over 2}$$
2. B.$${1 \over 16}$$
3. C.1
4. D.$${1 \over 4}$$