Jee Main
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
Jee Main
JEE Main 2026 (Online) 22nd January Morning Shift
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