MathematicsMedium38×since 2002Q3621Let x = x(y) be the solution of the differential equation 2y ex/y2dx+(y2−4xex/y2)dy=02y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 02yex/y2dx+(y2−4xex/y2)dy=0 such that x(1) = 0. Then, x(e) is equal to :Aeloge(2)e{\log _e}(2)eloge(2)B−eloge(2)- e{\log _e}(2)−eloge(2)Ce2loge(2){e^2}{\log _e}(2)e2loge(2)D−e2loge(2)- {e^2}{\log _e}(2)−e2loge(2)Check answerSkip