JEE Main 2021MathematicsDifferential EquationsMediumMCQ

JEE Main 2021 — Differential Equations Question with Solution

JEE Main 2021 (25 Jul Shift 2)

Question

Let y=yx be the solution of the differential equation xdy=y+x3cosxdx with yπ=0, then yπ2 is equal to:

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Show full solutionCorrect option: A
Correct answer
Aπ24+π2

Step-by-step explanation

We have, xdy=y+x3cosxdx

⇒xdy=ydx+x3cosxdx

⇒xdy-ydxx2=x3cosxdxx2

⇒ddxyx=∫xcosxdx

⇒yx=xsinx-∫1.sinxdx

Therefore, yx=xsinx+cosx+C

At x=π, y=0,

0=-1+C

⇒C=1,x=π,y=0

So, yx=xsinx+cosx+1

⇒y=x2sinx+xcosx+x

Hence, yπ2=π24+π2.

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About this question

This is a previous-year question from JEE Main 2021, 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.