JEE Main 2022 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 28 July 2022, Shift 1, Maths Q1

Question 1 of 90 in this shift, Maths question 1 of 30, Section A.

Let the solution curve of the differential equation

x\,dy=\left(\sqrt{x^2+y^2}+y\right)dx,\ x>0

, intersect the line

x=1

y=0

and the line

x=2

y=\alpha

. Then the value of

\alpha

is :

(1) $\frac{1}{2}$
(2) $\frac{3}{2}$ Official answer
(3) $-\frac{3}{2}$
(4) $\frac{5}{2}$

Official answer

Option 2

NTA final key (2022 Session 2).

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

Same idea in other shifts

Asked 5× in all

5 Apr 2026, Shift 1 · Q25Let $y = y(x)$ be the solution of the differential equation…HardNumerical value
9 Apr 2024, Shift 1 · Q10The solution of the differential equation $(x^2+y^2)dx-5xy\,dy=0$ , $y(1)=0$ , is :MediumSingle correct
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
9 Jan 2020, Shift 2 · Q64If $\frac{dy}{dx}=\frac{xy}{x^2+y^2}$ ; $y(1)=1$ ; then a value of $x$ satisfying $y(x)=e$ 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.