x^(y)^2+y^2*lnx+4=0,则dy/dx=?

y=lnx^2,dy/dx= 求解微分方程.dx/dy=x/[2(lnx-y)] y=x+lnx,求dx/dy 急.dy/dx-y/x=lnx 设y=(2+lnx)/x,则dy/dx(x=e)= y=x^2*4^x+lnx/cosx,求dy/dx 设y=x+lnx,则dx/dy= 解微分方程 y lny dx-x lnx dy=0 (x^2)dy+（y^2)dx=dx-dy dy/dx-y/x=x^2 y=[sin(x^4)]^2,则dy/dx=?,dy^2/dx^2=?,dy/d(x^2)=? 求微分方程dx/dy=x/2(lnx-y)的通解