Re: [理工] [工數]-BESSEL 標準..
※ 引述《mdpming (★ｐｉｇｍｉｎｇ★)》之銘言：
: 1.
: 2 2 1
: t z'' + tz' + (t - ---)z = 0
: 4
: 1 1
: z = c1 ------- cost + c2 ------- sint
: ----- -----
: \| t \| t
: 有高手可以提示一下嗎QQ
: 好多結果都是這樣@@
oo n+r
z = Σ an t
n=0
直接代入了
oo n+r oo n+r oo n+r+2 oo 1 n+r
Σ an(n+r-1)(n+r) t + Σ (n+r)an t + Σ an t -Σ ── an t
n=0 n=0 n=0 n=0 4
放出底 並且比較同次項係數
[(r-1)r+ r- 1/4]a0 =0 (r+ 1/2) (r- 1/2)a0 = 0
[(r)(1+r)+1+r- 1/4 ]a1 = 0 (r+ 3/2) (r+ 1/2) a1 =0
解a0可得指標根為 -1/2 1/2
a1=0 可得 a3 a5 a(2n+1)......都為0
a0
a2 = - ─────────
(2+r+1/2)(2+r-1/2)
(-1)^2 a0
a4 = ──────────────────
(4+r+1/2)(2+r+1/2)(4+r-1/2)(2+r-1/2)
(-1)^n a0
a(2n) = ────────────────────────── 循環式
(2n+r+1/2).....(2+r+1/2) (2+r-1/2).....(2n+r-1/2)
r= -1/2代入 循環式
(-1)^n a0 (-1)^n a0
a(2n)= ──────────────────── = ─────
(2n)......(2) (1)........(2n-1) (2n)!
oo (-1)^n 2n -1/2 -1/2
z1 =a0 Σ ──── t t 展回馬克勞林級數 = t cost
n=0 (2n)!
-1/2
已知一解為 t cost
-1/2 2 S- 1/t
z2 = t cost S tsec t e
-1/2
= t cost tant
-1/2
= t sint
-1/2 -1/2
z = c1 t cost + c2 t sint
或是直接寫成bessel型態
z = c1J (t) +c2 Y (t)
1/2 1/2
寫級數解大概要算10幾分鐘
寫成bessel只要10秒鐘
自己取捨吧= =
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