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Differential Equations question

2019 · 8 Apr · Shift 1 · Q38

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JEE Main
Mathematics
Differential Equations
2019 · 8 Apr · Shift 1 · Q38

Differential Equations question

2019 · 8 Apr · Shift 1 · Q38

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

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 :

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

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