JEE Main 2021MathematicsDifferential EquationsEasyMCQ

JEE Main 2021 — Differential Equations Question with Solution

JEE Main 2021 (20 Jul Shift 1)

Question

Let y=yx be the solution of the differential equation ex1-y2 dx+yxdy=0, y1=-1
Then the value of y32 is equal to:

Choose an option

Show full solutionCorrect option: B
Correct answer
B1-4e6

Step-by-step explanation

We have,

ex1-y2 dx+yxdy=0, y1=-1

⇒  ex1-y2 dx=-yxdy

⇒   ∫-y1-y2 dy=∫xexdx

⇒   12∫d1-y21-y2 dy=∫xexdx

⇒ 1-y2=exx-1+c

Now, at x=1,y=-1, then

⇒  0=0+c⇒c=0

∴  1-y2=exx-1

At x=3, then

1-y32=2e3

⇒  1-y32=4e6

⇒  y32=1-4e6

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About this question

This is a previous-year question from JEE Main 2021, 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.