Let y = y(x) be the solution of the differential equation, (x² + 1)²dy/dx + 2x(x² + 1)y = 1 such that y(0) = 0. If ay(1) = π 32, then the value of 'a' is:

Differential Equations: JEE Main 2019 Mathematics question

JEE Main 2019 · 8 Apr, Morning shift · Q3592

JEE Main 20198 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002

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 :

JEE Main 2019 · 8 Apr, Morning shift · Q3592

JEE Main 20198 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002

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 :