例4解方程y=1+(y)2 解 少y=p→y=p→ l=1+p → arctan p=x+C 即p=tan(x+c1) →y=tan(x+c1)x =-ln cos(x+Cu+C2解方程 2 y = 1+ ( y) 解 令y = p  y = p 2 1 p dx dp  = + dx p dp = +  2 1 1  arctan p = x + c 即 tan( ) 1 p = x + c   y = tan(x + c1 )dx1 2 = −lncos( x + c ) + c 例4