JEE Main 202126 Aug, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be a solution curve of the differential equation (y+1)tan2x dx+tanx dy+y dx=0(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0(y+1)tan2xdx+tanxdy+ydx=0, x∈(0,π2)x \in \left( {0,{\pi \over 2}} \right)x∈(0,2π). If limx→0+xy(x)=1\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1x→0+limxy(x)=1, then the value of y(π4)y\left( {{\pi \over 4}} \right)y(4π) is :AOption A. −π4- {\pi \over 4}−4πBOption B. π4−1{\pi \over 4} - 14π−1COption C. π4+1{\pi \over 4} + 14π+1DOption D. π4{\pi \over 4}4πCheckSkip · show solution
