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 :

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 :

${1 \over 2}$