JEE Main 202329 Jan, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y=f(x)y=f(x)y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=ey(x+1)dx-x^2dy=0,y(1)=ey(x+1)dx−x2dy=0,y(1)=e. Then limx→0+f(x)\mathop {\lim }\limits_{x \to {0^ + }} f(x)x→0+limf(x) is equal toAOption A. e2{e^2}e2BOption B. 0COption C. 1e2{1 \over {{e^2}}}e21DOption D. 1e{1 \over e}e1CheckSkip · show solution
