就是将y'当成未知数,解一元一次方程而已:
-sin(x+y) * (1+y')+y'=0
-sin(x+y)-y'sin(x+y)+y'=0
[1-sin(x+y)]y'=sin(x+y)
y'=sin(x+y)/[1-sin(x+y)]
cos(x+y)+y=1
-sin(x+y)*(x+y)'+y'=0
-sin(x+y)*(x'+y')+y'=0
-sin(x+y)*(1+y')+y'=0
-sin(x+y)-y'sin(x+y)+y'=0
[1-sin(x+y)]y'=sin(x+y)
y'=sin(x+y)/[1-sin(x+y)]