JEE Main 201911 Jan, Morning shiftMathematicsMediumDifferential Equations163× since 2002If y(x) is the solution of the differential equation dydx+(2x+1x)y=e−2x, x>0, {{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,dxdy+(x2x+1)y=e−2x,x>0, where y(1)=12e−2,y\left( 1 \right) = {1 \over 2}{e^{ - 2}},y(1)=21e−2, thenAOption A. y(log_e2) = log_e4BOption B. y(x) is decreasing in (0, 1)COption C. y(log_e2) = loge24{{{{\log }_e}2} \over 4}4loge2DOption D. y(x) is decreasing in (12,1)\left( {{1 \over 2},1} \right)(21,1)CheckSkip · show solution
