JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 9 January 2020, Shift 2, Maths Q64

Question 64 of 75 in this shift, Maths question 14 of 25, Section A.

\frac{dy}{dx}=\frac{xy}{x^2+y^2}

;

y(1)=1

; then a value of

x

satisfying

y(x)=e

is :

(1) $\sqrt2\,e$
(2) $\sqrt3\,e$ Official answer
(3) $\frac{1}{2}\sqrt3\,e$
(4) $\frac{e}{\sqrt2}$

Official answer

Option 2

NTA final key (Jan 2020).

Chapter: Differential Equations
Topic: Homogeneous Equations
Idea tested: Substitution y = vx

Same idea in other shifts

Asked 5× in all

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29 Jun 2022, Shift 1 · Q2Let the solution curve of the differential equation $x\frac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$ , $y(1) = 3$ be $y = y(x)$ . Then $y(2)$ is…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.