Jeeify
MCQMediumJEE Main 2026 · 21 January, Shift 1Linear Differential Equations

Mathematics Question from JEE Main 2026 · 21 January, Shift 1

Let y=y(x)y = y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan⁡−1x) dx=0(1+x^2)dy+(y-\tan^{-1}x) \, dx=0, y(0)=1y(0) = 1. Then the value of y(1)y(1) is:

  • A

    4eπ/4−π2−1\frac{4}{e^{\pi/4}} - \frac{\pi}{2} - 1

  • B

    2eπ/4+π4−1\frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1

  • C

    2eπ/4−π4−1\frac{2}{e^{\pi/4}} - \frac{\pi}{4} - 1

  • D

    4eπ/4+π2−1\frac{4}{e^{\pi/4}} + \frac{\pi}{2} - 1

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Sign in to see the solution

Practice more Linear Differential Equations questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Start free trialMore on Linear Differential Equations

Related questions