MathematicsMedium38×since 2002Q3570Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation (2xlogex)dydx+2y=3xlogex,x>0\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0(2xlogex)dxdy+2y=x3logex,x>0 and y(e−1)=0y\left(e^{-1}\right)=0y(e−1)=0. Then, y(e)y(e)y(e) is equal toA−3e-\frac{3}{\mathrm{e}}−e3B−32e-\frac{3}{2 \mathrm{e}}−2e3C−23e-\frac{2}{3 \mathrm{e}}−3e2D−2e-\frac{2}{\mathrm{e}}−e2Check answerSkip