更新1:
其實我問題應該改成 1/(1+x^2)^0.5怎麼積?
solve xy''=2(1+(y')^2)^0.5
回答 (1)
2010-01-03 9:18 pm
✔ 最佳答案
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
收錄日期: 2021-04-30 14:16:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100103000015KK01636