solve xy''=2(1+(y')^2)^0.5

y"/[1+(y')^2]^0.5 = 2/x
積分得 ln| y'+(1+y'^2)^0.5| = ln|cx^2|
y'+(1+y'^2)^0.5= cx^2
倒數: (1+y'^2)^0.5 -y'= 1/(cx^2)
上兩式相減除以2: y'= (cx^2)/2 - 1/(2cx^2)
積分得 y= (cx^3)/6+ 1/(2cx) + c'

2010-01-05 01:21:43 補充：
int 1/sqrt(1+x^2) dx, by sub. x=tant
=int sect dt = ln| tant + sect | +c = ln| x+sqrt(1+x^2) | + c