JEE Main 2018MathematicsDifferential EquationsHardMCQ

JEE Main 2018 — Differential Equations Question with Solution

JEE Main 2018 (15 Apr)

Question

Let y=yx be the solution of the differential equation dydx+2y=fx, where fx= 1,x∈0, 10,otherwise. If y(0)=0, then y 32 is

Choose an option

Show full solutionCorrect option: D
Correct answer
De2-12e3

Step-by-step explanation

dydx+2y=fx is a linear differential equation.

Integrating Factor =e∫2dx=e2x

Solution of the above equation is

y. e2x=∫fx.e2xdx+C

yx=e-2x∫0xfx.e2xdx+Ce-2x

y0=0⇒C=0

⇒yx=e-2x∫0xfx.e2xdx

∴y32=e-3∫01e2xdx+∫13/20.dx=e-32e2-1=e2-12e3

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About this question

This is a previous-year question from JEE Main 2018, 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.