JEE Main 2019MathematicsDifferential EquationsEasyMCQ

JEE Main 2019 — Differential Equations Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

If y=y(x) is the solution of the differential equation, xdydx+2y=x2 satisfying y1=1, then y12 is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D4916

Step-by-step explanation

Given differential equation can be written as

dydx+2yx=x , which is a linear differential equation.

If =e∫2xdx=e2 ln x=x2

Solution is

y.x2=∫x.x2dx

⇒y.x2=x44+c

Since, given curve passes through 1,1

⇒c=34

Hence, yx=x24+34x2.

So, y12=4916.

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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.