→ ntust661:我有做泰勒與遞回，參考看看 02/03 16:31
推 jersam89:屌暴了!感謝你 02/03 16:36
→ ntust661:我的屌爆了... 02/03 16:36
推 t5d:我笑了XD 02/03 16:41
y※ 引述《jersam89 (123)》之銘言：
: y''+ 2y'+(1+x^2)y = 0
: 這該怎麼算?
: 我用自變數變更也求不出來耶~
: 謝謝嚕!
級數解= =
∞ n-2 ∞ n-1 ∞ n ∞ n+2
Σ n(n-1)an x + 2Σ n an x + Σ an x + Σ an x = 0
2 1 0 0
∞ n+2 ∞ n+2 ∞ n+2 ∞ n+2
Σ(n+4)(n+3)an+4 x + 2Σ(n+3) an+3 x + Σ an+2 x + Σ an x = 0
-2 -2 -2 0
2 a2 + 2 a1 + a0 + 6a3 x + 4a2 x + a1 x = 0
a2 = -a1 - a0/2
a3 = -2a2/3 - a1/6 = -2/3(-a1 - a0/2) - a1/6 = 1/2a1 + a0/3
Recurrence Relation
-2(n+3)an+3 - an+2 -an
an+4 = ─────────────────────
(n+4)(n+3)
-6a3 - a2 -a0
a4 = ───────
12
-8a4 -a3 -a1
a5 = ──────
20
2 3
x x 2 1 3
y = a0(1 - ── + ── + ...) + a1(x - x + ── x + ...)
2 3 2
--
Taylor
y'' = -2y'- (1+x^2)y
y''(0) = -2y' - y
y'''= -2y'' - [(1+x^2)y' +2xy]
y'''(0) = -2y'' - y' = -2(-2y'-y) - y' = 3y'+2y
y^[4]= -2y''' -[2y + 2*(2x)y' + (1+x^2)y'']
y^[4](0)= -2y''' - 2y - y'' = -4y'-5y
a0 a1 a2 a3 a4
y = ── + ── + ── + ── + ── + ...
0! 1! 2! 3! 4!
x^2 x^3 5x^4 2 x^3 x^4
= a0(1 - ── + ── - ──) + a1(x - x + ── - ── + ...)
2 3 4! 2 6
能力只到此了= =...
--
有錯請指正唷!!
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