JEE Main 2020MathematicsDifferential EquationsMediumMCQ

JEE Main 2020 — Differential Equations Question with Solution

JEE Main 2020 (05 Sep Shift 1)

Question

If y=yx is the solution of the differential equation 5+ex2+y⋅dydx+ex=0 satisfying y0=1 then value of y(loge13) is

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Show full solutionCorrect option: B
Correct answer
B-1

Step-by-step explanation

Given dy2+y=-exdx5+ex

⇒∫dy2+y=-∫exdx5+ex

⇒loge(2+y)=-loge5+ex+logeC

⇒loge(2+y)=logeC5+ex

⇒y=C5+ex-2

∵y(0)=1

∴ c=18

y=185+ex-2

y(loge13)=185+eloge13-2

⇒y(loge13)=185+13-2

∴  yloge13=-1

 

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About this question

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.