JEE Main 2019MathematicsDifferential EquationsMediumMCQ

JEE Main 2019 — Differential Equations Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

Let y=y(x) be the solution of the differential equation, x2+12 dydx+2x(x2+1)y=1 such that y0=0. If a y1=π32, then the value of a is

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Show full solutionCorrect option: A
Correct answer
A116

Step-by-step explanation

x2+12 dydx+2xx2+1y=1
⇒dydx+2xx2+1y=1(x2+1)2
Which is a linear differential equation with integrating factor
I.F. = e∫2xx2+1dx=eln(x2+1)=x2+1
The solution of the given differential equation will be,
y.x2+1=∫x2+1.1x2+12dx=∫dxx2+1=tan-1x+c
Now, y0= 0 ⇒C=0 and y=1(x2+1)tan-1x
ay(1)=π32⇒a=π/32π/8=14
⇒a=116

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About this question

This is a previous-year question from JEE Main 2019, 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.