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

Let y = y(x) be the solution of the differential equation, \({\rm{}}{

1.	Let y = y(x) be the solution of the differential equation, \({\rm{}}{\left( {{{\rm{x}}^2} + 1} \right)^2}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + 2{\rm{x}}\left( {{{\rm{x}}^2} + 1} \right){\rm{y}} = 1\) is such that y(0) = 0. If \(\sqrt {\rm{a}} {\rm{\;y}}\left( 1 \right) = \frac{{\rm{\pi }}}{{32}},{\rm{\;then\;the}}\) value of ‘a’ is:
A.	\(\frac{1}{4}\)
B.	\(\frac{1}{2}\)
C.	1
D.	\(\frac{1}{{16}}\)
Answer» E.

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