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Differential Equations

JEE Main 2026 (Online) 22nd January Morning Shift

MCQ+4 / -12026

Let the solution curve of the differential equation $x d y-y d x=\sqrt{x^2+y^2} d x, x>0$, $y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to


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