Jee Main
Differential Equations
JEE Main 2026 (Online) 21st January Morning Shift
MCQ+4 / -12026
Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^2\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1$. Then the value of $y(1)$ is :
