JEE Main 2017MathematicsDifferential EquationsMediumMCQ

JEE Main 2017 — Differential Equations Question with Solution

JEE Main 2017 (02 Apr)

Question

If 2+sinxdydx+y+1cosx=0 and y0=1, then yπ2 is equal to

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Show full solutionCorrect option: A
Correct answer
A13

Step-by-step explanation

We have,

2+sinxdydx=-y+1cosx

⇒∫dyy+1=-∫cosxsinx+2dx

Let, sinx=t⇒cosxdx=dt

⇒logy+1=-∫dtt+2+C

where, C is the constant of integration.

⇒logy+1+logsinx+2-C=0

⇒logy+1sinx+2=C

Given, y0=1

⇒C=log4.

⇒logy+1sinx+2=log4    .....i

Now, put x=π2 in the equation i, we get, logy+1+log3-log4=0

⇒ logy+1=log43

⇒y+1=43

⇒y=43-1=13.

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About this question

This is a previous-year question from JEE Main 2017, 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.