Let $y = y(x)$ be a solution curve of the differential equation. $(1 - x^2y^2)\,dx = y\,dx + x\,dy$. If the line $x = 1$ intersects the curve $y = y(x)$ at $y = 2$ and the line $x = 2$ intersects the curve $y = y(x)$ at $y = \alpha$, then a value of $\alpha$ is

Option 3: $\frac{1+3e^2}{2(3e^2-1)}$

JEE Main 2023 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 11 April 2023, Shift 1, Maths Q20

Question 20 of 90 in this shift, Maths question 20 of 30, Section A.

Let

y = y(x)

be a solution curve of the differential equation.

(1 - x^2y^2)\,dx = y\,dx + x\,dy

. If the line

x = 1

intersects the curve

y = y(x)

y = 2

and the line

x = 2

intersects the curve

y = y(x)

y = \alpha

, then a value of

\alpha

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

Official answer

Option 3

NTA final key.

Chapter: Differential Equations
Topic: Variable Separable Form
Idea tested: Substitution of Product (u = xy)

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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.