JEE Main 20246 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (1+x2)dydx+y=etan−1x\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}(1+x2)dxdy+y=etan−1x, y(1)=0y(1)=0y(1)=0. Then y(0)y(0)y(0) isAOption A. 14(eπ/2−1)\frac{1}{4}\left(e^{\pi / 2}-1\right)41(eπ/2−1)BOption B. 12(1−eπ/2)\frac{1}{2}\left(1-e^{\pi / 2}\right)21(1−eπ/2)COption C. 14(1−eπ/2)\frac{1}{4}\left(1-e^{\pi / 2}\right)41(1−eπ/2)DOption D. 12(eπ/2−1)\frac{1}{2}\left(e^{\pi / 2}-1\right)21(eπ/2−1)CheckSkip · show solution
