JEE Main 2024MathematicsDifferential EquationsMediumMCQ

JEE Main 2024 — Differential Equations Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

If sinyx=loge|x|+α2 is the solution of the differential equation xcosyxdydx=ycosyx+x and y(1)=π3, then α2 is equal to

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Show full solutionCorrect option: A
Correct answer
A3

Step-by-step explanation

Given: xcosyxdydx=ycosyx+x

⇒cosyxdydx=yxcosyx+1

Putting, y=vx

⇒dydx=v+xdvdx

⇒cosvv+xdvdx=vcosv+1

⇒v+xdvdx=v+1cosv

⇒xdvdx=1cosv

⇒cosvdv=dxx

⇒∫cosvdv=∫dxx

⇒sinv=logx+c

⇒sinyx=logx+c

Now, f1=π3

⇒sinπ3=log1+c

⇒32=c

⇒sinyx=logx+32

So, on comparing we get,

⇒α=3

⇒α2=3

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About this question

This is a previous-year question from JEE Main 2024, 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.