JEE Main 2019 — Differential Equations Question with Solution
From: JEE Main 2019 (Online) 12th January Morning Slot
Question
Let y = y(x) be the solution of the differential equation, x + y = x loge x, (x > 1). If 2y(2) = loge 4 1, then y(e) is equal to :
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Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
for
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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.