JEE Main 2018MathematicsDifferential EquationsHardMCQ

JEE Main 2018 — Differential Equations Question with Solution

JEE Main 2018 (08 Apr)

Question

Let y=yx be the solution of the differential equation sin⁡xdydx+ycosx⁡=4x, x∈0, π. If yπ2=0, then yπ6 is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D-89π2

Step-by-step explanation

dydx+ycot x=4xsinx, x∈0, π

I.F. =e∫cotx=eln⁡sinx=sinx

∴y sinx=∫4xsinx.sinx dx

ysinx=2x2+c

∵yπ2=0

∴0×1=2×π22+c

⇒c=-π22

yπ6=2×π62-π2212=π218-π22×2

=π29-π2

=-8π29

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About this question

This is a previous-year question from JEE Main 2018, 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.