JEE Main 2022MathematicsDifferential EquationsHardNumerical

JEE Main 2022 — Differential Equations Question with Solution

JEE Main 2022 (27 Jun Shift 2)

Question

Let y=yx be the solution of the differential equation 1-x2dy=xy+x3+21-x2dx,-1<x<1
and y0=0. If ∫-12121-x2yxdx=k then k-1 is equal to

Enter your answer

Show full solutionCorrect answer: 320
Correct answer
320

Step-by-step explanation

Given,

1-x2dydx=xy+x3+21-x2

⇒dydx+-x1-x2y=x3+21-x2

IF=e∫-x1-x2dx=1-x2

yx·1-x2=x44+2x+c

y0=0⇒c=0

1-x2yx=x44+2x

So,required value =∫-1212x44+2xdx-14·2∫012x4dx

=110x5012=1320

k-1=320

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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.