JEE Main 2015MathematicsDifferential EquationsMediumMCQ

JEE Main 2015 — Differential Equations Question with Solution

JEE Main 2015 (11 Apr Online)

Question

The solution of the differential equation ydx-x+2y2dy=0 is x=fy. If f-1=1, then f(1) is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B3

Step-by-step explanation

ydx-xdy=2y2dy

ydx-xdyy2=2dy

dxy=2dy

∫dxy=2∫dy

xy=2y+c

x=2y2+cy

If y= -1, x=1

1=2-c

c=1

∴ Curve would be

x=2y2+y

If y=1 then x=3 .

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About this question

This is a previous-year question from JEE Main 2015, 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.