JEE Main 2023MathematicsDifferential EquationsMediumMCQ

JEE Main 2023 — Differential Equations Question with Solution

JEE Main 2023 (29 Jan Shift 1)

Question

Let y=f(x) be the solution of the differential equation y(x+1)dx-x2dy=0, y(1)=e. Then limx→0+f(x) is equal to

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Show full solutionCorrect option: A
Correct answer
A0

Step-by-step explanation

Given,

y(x+1)dx-x2dy=0

⇒x+1x2dx=dyy

⇒1x+1x2dx=dyy

Now integrating both side we get,

logex-1x=logey+c

Now on using y1=e we get, c=-2

So, the equation of curve becomes logex-1x=logey-2

⇒y=elnx-1x+2

Hence, limx→0+elnx-1-1x+2=e-∞=0.

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About this question

This is a previous-year question from JEE Main 2023, 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.