Re: [理工] [工數]-高階ODE
看板Grad-ProbAsk作者ntust661 (Auf Wiedersehen!)時間14年前 (2010/02/21 00:19)推噓3(3推 0噓 4→)留言7則, 4人參與討論串36/46 (看更多)
※ 引述《arroyo0301 (arroyo)》之銘言：
: (a)(5%)Define a linear operator L3 = D^3 - 6D^2 + 12D - 8.
: Please determine the complementary function yc(x), a particular
: solution yp(x) and the general solution for L3(y) = 4e^(2x)
: (b)(5%)In the yc(x), if the e^2x term is removed, please find the
: second-order linear differential equation L2(y) = 0 whose
: complementary function is the linear combination of the left terms
: and coefficient of D^2 is 1.
: A小題沒有問題 但是B小題感覺怪怪的 題目的意思是找出一個解為
: xe^2x x^2e^2x的2階ODE嗎@@ 我用y` y``時再湊不出答案 希望高手幫忙
凡阿!!
直接第二題
2x 2 2x
y = c1 x e + c2 x e
2x
x e 為 ODE 中的齊性解
配 ODE
2x
y = c1 x e
2x 2x
y'= c1 e + 2 c1 x e
2x 2 2x
xy' = c1 x e + 2 c1 x e
2 2x
xy = c1 x e
xy' - 2xy - y = 0
xy' - (2x + 1)y = 0
2x + 1
y' - ────── y = 0
x
2x + 1
(D - P(x))(D - ────) y = 0
x
代入第二齊性解
2 2x
y = x e
2x 2 2x
y' = 2x e + 2 x e
2x
y/x = x e
y' - 2y - 2y/x = 0
1
y' - 2(1 + ──) y = 0
x
2
P(x) = ( 2 + ──)
x
2 1
(D - (2 + ───)) (D - (2 + ───)) = 0
x x
展開，
2 1 1 2 1
D - D(2 + ──) - (2 + ──)D + (2 + ──)(2 - ──) = 0
x x x x
2 1 2 2 1
{D - (2 + ──)D + (4 + ── - ──) + ── } y = 0
x x x^2 x^2
得解!
1 2 1
y'' - (2 + ──) y' + (4 + ── - ──) y = 0
x x x^2
希望沒錯QQ
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