JEE Main 2024MathematicsDifferential EquationsEasyNumerical

JEE Main 2024 — Differential Equations Question with Solution

JEE Main 2024 (01 Feb Shift 2)

Question

If dxdy=1+x−y2y,x1=1, then 5x2 is equal to:

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Show full solutionCorrect answer: 5
Correct answer
5

Step-by-step explanation

Given: dxdy=1+x-y2y

⇒dxdy=1+xy-y

⇒dxdy-1+xy=-y

Now let 1+x=t⇒dxdy=dtdy

⇒dtdy-ty=-y

⇒IF=e∫-1ydy

⇒IF=e-logy

⇒IF=elog1y

⇒IF=1y

So, the solution of the given differential equation is

ty=∫1y×-ydy

⇒1+xy=∫1y×-ydy

⇒1+xy=-y+c

Also, x1=1

⇒21=-1+c

⇒c=3

⇒1+xy=-y+3

Putting y=2

⇒1+x2=-2+3

⇒1+x=2

⇒x=1

⇒5x=5

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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.