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Question #41475

Difficulty: 8/10 | 🟢 Free

The Question

Let y=y(x)y = y(x) be the continuously differentiable solution to the differential equation xdydx=y+x2−y2,x≥1x \frac{dy}{dx} = y + \sqrt{x^2 - y^2}, \quad x \ge 1 satisfying the initial condition y(1)=0y(1) = 0. Find the value of the integral ∫1eπy(x) dx\int_{1}^{e^\pi} y(x) \, dx.

Options

  • A.

    5e2π−eπ+210\frac{5e^{2\pi} - e^\pi + 2}{10}

  • B.

    e2π+15\frac{e^{2\pi} + 1}{5}

  • C.

    5e2π+eπ−210\frac{5e^{2\pi} + e^\pi - 2}{10}

  • D.

    2e2π−eπ+15\frac{2e^{2\pi} - e^\pi + 1}{5}

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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

The Sign-Consistency Principle|Homogeneous Function Homogeneity|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' and the 'Sign-Consistency Principle' in the context of homogeneous differential equations.

When students substitute y=vxy = vx, they obtain xv′=1−v2x v' = \sqrt{1-v^2}.

Dividing by 1−v2\sqrt{1-v^2} implicitly assumes v≠1v \neq 1, which masks the singular solution y=xy=x.

Furthermore, integrating blindly yields y=xsin⁡(ln⁡x)y = x \sin(\ln x) for all xx, but substituting this back into the original DE reveals a sign inconsistency for x>eπ/2x > e^{\pi/2} because x2−y2\sqrt{x^2-y^2} must be non-negative, forcing cos⁡(ln⁡x)≥0\cos(\ln x) \ge 0.

Recognizing that v(x)v(x) reaches its maximum domain limit at x=eπ/2x = e^{\pi/2} and must smoothly transition to the singular branch y=xy=x is a hallmark of deep conceptual mastery required for JEE Advanced.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Integrals - Class 12
ConceptThe Sign-Consistency Principle
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree