Official solution
\[{({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1\]
\[ \Rightarrow \] \[{{dy} \over {dx}} + \left( {{{2x} \over {1 + {x^2}}}} \right)y = {1 \over {{{\left( {1 + {x^2}} \right)}^2}}}\]
I.F = \[{e^{\int {{{2x} \over {1 + {x^2}}}dx} }} = {e^{\ln \left( {1 + {x^2}} \right)}} = 1 + {x^2}\]
\[ \therefore \] \[{d \over {dx}}\left( {\left( {1 + {x^2}} \right)y} \right) = {1 \over {1 + {x^2}}}\]
Integrating both sides we get,
\[{\left( {1 + {x^2}} \right)y}\] = \[{\tan ^{ - 1}}x\] + C
When x = 0 then y = 0
So, 0 = 0 + C
\[ \Rightarrow \] C = 0
\[ \Rightarrow \] \[{\left( {1 + {x^2}} \right)y}\] = \[{\tan ^{ - 1}}x\]
Put x = 1,
y.(2) = \[{\pi \over 4}\]
\[ \Rightarrow \] y = \[{\pi \over 8}\] = y(1)
Given,
\[\sqrt ay(1)\] = \[\pi \over 32\]
\[ \Rightarrow \] \[\sqrt a. {\pi \over 8}\] = \[\pi \over 32\]
\[ \Rightarrow \] \[\sqrt a = {1 \over 4}\]
\[ \Rightarrow \] \[a = {1 \over {16}}\]