JEE Main 202621 Jan, Morning shiftMathematicsEasyDifferential Equations163× since 2002Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan−1x) dx=0(1+x^2)dy+(y-\tan^{-1}x)\,dx=0(1+x2)dy+(y−tan−1x)dx=0, y(0)=1y(0)=1y(0)=1. Then the value of y(1)y(1)y(1) is :AOption A. 4eπ/4−π2−1\frac{4}{e^{\pi/4}}-\frac{\pi}{2}-1eπ/44−2π−1BOption B. 2eπ/4−π4−1\frac{2}{e^{\pi/4}}-\frac{\pi}{4}-1eπ/42−4π−1COption C. 4eπ/4+π2−1\frac{4}{e^{\pi/4}}+\frac{\pi}{2}-1eπ/44+2π−1DOption D. 2eπ/4+π4−1\frac{2}{e^{\pi/4}}+\frac{\pi}{4}-1eπ/42+4π−1CheckSkip · show solution
