JEE Main 2020MathematicsDifferential EquationsMediumMCQ

JEE Main 2020 — Differential Equations Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

Let y=y(x) be the solution of the differential equation, xy'-y=x2(xcosx+sinx),x>0. If y(π)=π, then y''π2+yπ2 is equal to :

Choose an option

Show full solutionCorrect option: A
Correct answer
A2+π2

Step-by-step explanation

Given xdydx−y=x2xcos+sinx

⇒dydx−1xy=xxcosx+sinx

∴ I.F =e-lnx=1x

∴ Solution is y.1x=∫1x.xxcosx+sinxdx

yx=∫xcosx+sinxdx

yx=xsinx+C

∵yπ=π ⇒C=1

y=x2sinx+x

dydx=x2cosx+2xsinx+1

d2ydx2=-x2sinx+2xcosx+2sinx+2xcosx

=−x2sinx+4xcosx+2sinx

∴ y''π2+yπ2=−π24+0+2+π24+π2

=−π24+2+π24+π2

=π2+2.

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About this question

This is a previous-year question from JEE Main 2020, covering the Differential Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.