$${({x^2} + 1)^2}{{dy} \over {dx}} + 2x({x^2} + 1)y = 1$$

$$ \Rightarrow $$ $${{dy} \over {dx}} + \left( {{{2x} \over {1 + {x^2}}}} \right)y = {1 \over {{{\left( {1 + {x^2}} \right)}^2}}}$$

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,

$$\sqrt ay(1)$$ = $$\pi \over 32$$

$$ \Rightarrow $$ $$\sqrt a. {\pi \over 8}$$ = $$\pi \over 32$$

$$ \Rightarrow $$ $$\sqrt a = {1 \over 4}$$

$$ \Rightarrow $$ $$a = {1 \over {16}}$$