JEE Main 2021MathematicsDifferential EquationsEasyMCQ

JEE Main 2021 — Differential Equations Question with Solution

JEE Main 2021 (16 Mar Shift 2)

Question

If y=yx is the solution of the differential equation dydx+tanxy=sinx,0≤x≤π3, with y0=0, then yπ4 is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B

122loge2

Step-by-step explanation

dydx+tanxy=sinx; 0≤x≤π3

I.F.=e∫tanxdx=elnsecx=secx

ysecx=∫tanxdx+C

ysecx=lnsecx+C

x=0, y=0  ⇒ C=0

ysecx=lnsecx

y=cosx·lnsecx

yx=π4=12·ln2

yx=π4=122loge2

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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.