解方程组y－2＝0,x＋y－1＝0得x＝－1,y＝2,作变换X＝x＋1,Y＝y－2,则原微分方程化为dY/dX＝2Y^2/(X＋Y)^2＝2(Y/X)/[1＋(Y/X)]^2令u＝Y/X,则u＋du/dX＝2u^2/(1+u)^2,分离变量得(1+u)^2/[u(u^2+1)]du＝－dX/X[1/u＋2/(...