另解一y=p(x),ym=p(x),p'=√1+p2, ≤dx,则P+√1+p2)=x+hnC1, 1+ p+√1+p2=Ce, 1+ 2 e e 1+ P e 2 e C 2 (C1 e ee (C )+C2 2 e 十(2+C3 K心另解一 y = p(x), y = p(x), 1 , 2 p = + p , 1 2 dx p dp = + ln( 1 ) ln , 1 2 p + + p = x + C 1 , 1 2 x p + + p = C e , 1 1 1 2 x C e p p = + − 1 , 1 2 C e p p − x + − = ( ), 2 1 1 1 C e p C e x x − = − ( ), 2 1 1 1 C e y C e x x −  = −  −  = − dx C e y C e x x ( ) 2 1 1 1 2 1 1 ( ) 2 1 C C e C e x x = + + − ( ) . 2 1 2 3 1 1 C x C C e y C e x x = − + + −