MCQMediumJEE Main 2026 · 22 January, Shift 1Homogeneous Differential Equations

Mathematics Question from JEE Main 2026 · 22 January, Shift 1

Let the solution curve of the differential equation xdyydx=x2+y2dx,x>0,x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx,\quad x>0, with y(1)=0y(1)=0, be y=y(x)y=y(x). Then y(3)y(3) is equal to

  • A

    44

  • B

    22

  • C

    11

  • D

    66

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