Differential Equations: JEE Main 2020 Mathematics question

JEE Main 2020 · 4 Sep, Morning shift · Q3597

JEE Main 20204 Sep, Morning shiftMathematicsMediumDifferential Equations163× since 2002

Let y = y(x) be the solution of the differential equation, xy'- y = x²(xcosx + sinx), x > 0. if y (

\pi

) =

\pi

then

y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)

is equal to :

JEE Main 2020 · 4 Sep, Morning shift · Q3597

JEE Main 20204 Sep, Morning shiftMathematicsMediumDifferential Equations163× since 2002

Let y = y(x) be the solution of the differential equation, xy'- y = x²(xcosx + sinx), x > 0. if y (

\pi

) =

\pi

then

y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)

is equal to :

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