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Question #41600

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = y(x)$ be the solution to the differential equation $x \frac{dy}{dx} = y + x \sqrt{1 - \sin\left(\frac{y}{x}\right)}, \quad x > 0$ with the initial condition $y(1) = \pi$ . If $y(x_0) = \frac{3\pi}{2} x_0$ for some $x_0 > 1$ , then the value of $x_0$ is:

Options

A.
$(\sqrt{2} + 1)^{\sqrt{2}}$
B.
$(\sqrt{2} - 1)^{\sqrt{2}}$
C.
$(\sqrt{2} + 1)^{\frac{1}{\sqrt{2}}}$
D.
$(\sqrt{2} - 1)^{\frac{1}{\sqrt{2}}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Substitution Logic in Homogeneous DE|Trigonometric Identical Substitutions|ASTC Quadrant Sign Convention

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle domain-restricted trigonometric identities within the context of a homogeneous differential equation.

The core trap lies in the simplification of $\sqrt{1-\sin v}$ .

Students habitually write $\sqrt{1-\sin v} = \cos(v/2) - \sin(v/2)$ without checking the domain.

Because the initial condition $y(1)=\pi$ places $v$ in the second quadrant (and moving to the third), $\sin(v/2) > \cos(v/2)$ , making the correct branch $\sin(v/2) - \cos(v/2)$ .

Failing to apply the ASTC quadrant sign convention leads to a negated integral and the reciprocal trap answer $(\sqrt{2}-1)^{\sqrt{2}}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Differential Equations - Class 12
Concept	Substitution Logic in Homogeneous DE
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free