JEE Main 2022MathematicsDifferential EquationsMediumMCQ

JEE Main 2022 — Differential Equations Question with Solution

JEE Main 2022 (28 Jun Shift 2)

Question

Let x=xy be the solution of the differential equation 2yexy2dx+y2-4xexy2dy=0 such that x1=0. Then, xe is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D-e2loge2

Step-by-step explanation

Given,

2yexy2dx+y2-4xexy2dy=0

2exy2ydx-2xdy+y2dy=0

2exy2y2dx-x·2ydyy+y2dy=0

Divide by y3

2exy2y2dx-x·2ydyy4+1ydy=0

2exy2dxy2+1ydy=0

Now integrating both side we get,

∫2exy2dxy2+∫1ydy=0

2exy2+lny+c=0

Given, 0,1 lies on it, 

So, 2e0+ln1+c=0⇒c=-2 

Hence required curve: 2exy2+lny-2=0

For xe

2exe2+lne-2=0  ⇒ x=-e2loge2

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About this question

This is a previous-year question from JEE Main 2022, 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.