JEE Main 202226 Jun, Evening shiftMathematicsMediumDifferential Equations163× since 2002If y=y(x)y = y(x)y=y(x) is the solution of the differential equation xdydx+2y=x exx{{dy} \over {dx}} + 2y = x\,{e^x}xdxdy+2y=xex, y(1)=0y(1) = 0y(1)=0 then the local maximum value of the function z(x)=x2y(x)−ex, x∈Rz(x) = {x^2}y(x) - {e^x},\,x \in Rz(x)=x2y(x)−ex,x∈R is :AOption A. 1 −-− eBOption B. 0COption C. 12{1 \over 2}21DOption D. 4e−e{4 \over e} - ee4−eCheckSkip · show solution
