JEE Main 2026 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2026, Shift 2, Maths Q20

Question 20 of 75 in this shift, Maths question 20 of 25, Section A.

Let

x = x(y)

be the solution of the differential equation

2y^2\frac{dx}{dy} - 2xy + x^2 = 0, y > 1, x(e) = e

. Then

x(e^2)

is equal to:

(1) $\frac{3}{2}e^2$
(2) $\frac{2}{3}e^2$ Official answer
(3) $e^2$
(4) $2e^2$

Official answer

Option 2

NTA final key.

Chapter: Differential Equations
Topic: Homogeneous Equations
Idea tested: Identifying Homogeneous Functions

Same topic in other shifts

All Homogeneous Equations questions

5 Apr 2026, Shift 1 · Q25Let $y = y(x)$ be the solution of the differential equation…HardNumerical value
6 Apr 2024, Shift 2 · Q13Suppose the solution of the differential equation $\frac{dy}{dx}=\frac{(2+\alpha)x-\beta y+2}{\beta x-2\alpha y-(\beta\gamma-4\alpha)}$ …HardSingle correct
8 Apr 2024, Shift 2 · Q27Let $\alpha|x| = |y|e^{xy-\beta}$ , $\alpha, \beta \in \mathbb{N}$ 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
31 Jan 2024, Shift 1 · Q12The solution curve of the differential equation $y\frac{dx}{dy}=x(\log_ex-\log_ey+1),\ x>0,\ y>0$ passing through the point $(e,1)$ isMediumSingle correct
10 Apr 2023, Shift 1 · Q10The slope of tangent at any point $(x,y)$ on a curve $y=y(x)$ is $\frac{x^2+y^2}{2xy}$ , $x>0$ . If $y(2)=0$ , then a value of $y(8)$ isMediumSingle 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 108 of the 174 JEE Main shifts held from 2017 to 2026.