JEE Main 2020 — Differential Equations Question with Solution
From: JEE Main 2020 (Online) 5th September Evening Slot
Question
Let y = y(x) be the solution of the differential
equation
cosx + 2ysinx = sin2x, x .
If y = 0, then y is equal to :
cosx + 2ysinx = sin2x, x .
If y = 0, then y is equal to :
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
cosx + 2ysinx = sin2x
I.F = = sec2 x
y.sec2 x =
ysec2x =
ysec2x = 2secx + c
Given at x = , y = 0
0 =
c = -4
ysec2x = 2secx - 4
Here put x =
y.2 = - 4
y =
I.F = = sec2 x
y.sec2 x =
ysec2x =
ysec2x = 2secx + c
Given at x = , y = 0
0 =
c = -4
ysec2x = 2secx - 4
Here put x =
y.2 = - 4
y =
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This is a previous-year question from JEE Main 2020, 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.