换元u=y-x,则dy/dx=du/dx+1,原方程化为du/dx=-cosu,分离变量du/cosu=-dx,两边积分ln(secu+tanu)=-lnx+lnC,所以x(secu+tanu)=C,代入u=y-x得通解x(sec(y-x)+tan(y-x))=C