Jeeify
MCQMediumJEE Main 2025 · 7 April, Shift 1Linear Differential Equations

Mathematics Question from JEE Main 2025 · 7 April, Shift 1

Let y=y(x)y = y(x) be the solution curve of the differential equation x(x2+ex) dy+(ex(x−2)y−x3) dx=0,x>0, x(x^2 + e^x) \, dy + \left( e^x(x - 2) y - x^3 \right) \, dx = 0, \quad x > 0, passing through the point (1,0)(1, 0). Then y(2)y(2) is equal to:

  • A

    44−e2\frac{4}{4 - e^2}

  • B

    22+e2\frac{2}{2 + e^2}

  • C

    22−e2\frac{2}{2 - e^2}

  • D

    44+e2\frac{4}{4 + e^2}

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