QJEEM19Maea73: JEE Main 2019 Mathematics - Class 12 JEEM19Maea73 Solution: Let y = y(x) be the solution of the differential equation, such …

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

{({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}}

Step-by-Step Verified Solution