y=y(log y-log x+1)," then the solution of the equation is "

If x(dy)/(dx)=y(log y -logx+1), then the solution of the equation is

IF y' = y/x(log y - log x+1), then the solution of the equation is :

IF y' = y/x(log y - log x+1), then the solution of the equation is :

If x(dy)/(dx) = y (log y - log x +1) , then the solution of the equation is

If x(dy)/(dx)=y(log_(e)y-log_(e)x+1) ,then solution of the equation is

If x dy/dx=y(logy-logx+1) , then the solution of the differential equation is (A) log(x/y)=Cy (B) log(y/x)=Cy (C) log(x/y)=Cx (D) log(y/x)=Cx

Let y(x) be the solution of the differential equation . (x log x) (dy)/(dx) + y = 2x log x, (x ge 1) . Then y (e) is equal to