JEE Main 2018 — Differential Equations Question with Solution
From: JEE Main 2018 (Online) 15th April Evening Slot
Question
The curve satifying the differeial equation, (x2 y2) dx + 2xydy = 0 and passing through the point (1, 1) is :
Choose an option
Show full solutionCorrect option: A
Correct answer
Aa circle of radius one.
Step-by-step explanation
(x2 y2) dx + 2xydy = 0
=
Let y = vx
= v + x
v + x = v + x =
x =
=
After intergrating, we get
= ln + lnc
+ 1 =
As curve passes through the point (1, 1), so 1 + 1 = c
c = 2
x2 + y2 2x = 0, which is a circle of radius one.
=
Let y = vx
= v + x
v + x = v + x =
x =
=
After intergrating, we get
= ln + lnc
+ 1 =
As curve passes through the point (1, 1), so 1 + 1 = c
c = 2
x2 + y2 2x = 0, which is a circle of radius one.
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This is a previous-year question from JEE Main 2018, 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.