JEE Main 2023MathematicsDifferential EquationsEasyMCQ

JEE Main 2023 — Differential Equations Question with Solution

JEE Main 2023 (10 Apr Shift 1)

Question

The slope of tangent at any point x, y on a curve y=yx is x2+y22xy, x>0. If y2=0, then a value of y8 is

Choose an option

Show full solutionCorrect option: D
Correct answer
D43

Step-by-step explanation

Given:

dydx=x2+y22xy

Put y=vx⇒dydx=v+xdvdx

v+xdvdx=1+v22v

⇒xdvdx=1-v22v

⇒∫2vv2-1dv=-∫dxx

⇒logev2-1=logeCx

⇒y2-x2x2=Cx

⇒y2-x2=Cx

Put x=2 and y=0 we get,

0-22=2C⇒C=-2

⇒y2=x2-2x

⇒y8=82-16

⇒y8=48=43

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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.