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Anonymous asked in 科學數學 · 1 decade ago

Second-Order Linear ODEs的相關問題

2.6 Existence and Uniqueness of Solutions. Wronskian

Bases of solutions. Corresponding ODEs. Wronskians

Find an ODE (1) for which the given functions are solutions. Show linear independence (a) by considering quotients, (b) by Theorem 2.

1. e^kx,xe^kx

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  • 1 decade ago
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    1.方程式

    輔助方程式有3根, 0, k, k => (m-0)(m-k)(m-k)=0

    => m³-2km²+k²m=0

    故原線性ODE為 y'''- 2ky" + ky' =0

    2.線性獨立(此題應設k不為0,否則題目錯誤)

    (法一)比例法

    e^(kx), xe^(kx) 無任何兩個函數比例(quotient)為常數現象,故為線性獨立 (註:此法不嚴謹, 如 1, x+1, 3-2x, 三個函數亦無倍數關係,但事實上此三個函數非線性獨立)

    (法二)解方程式法(定義)

    設 a + b e^(kx)+ c xe^(kx) = 0 (欲求 a, b, c是否有非零者)

    設x=0 => a+b=0 ---(A)

    一階導函數,再令x=0 => bk + c = 0 --(B)

    二階導函數,再令x=0 => bk² + 2k c =0 --(C)

    解(A),(B),(C)聯立,則 a=0, b=0, c=0

    由定義知 1, e^(kx), xe^(kx) 為線性獨立

    (法三) Wronskian 法

    W(1, e^(kx), xe^(kx) )

    =| 1 e^(kx) xe^(kx) |

    | 0 ke^(kx) (kx+1)e^(kx) |

    | 0 k²e^(kx) (k²x+2k)e^(kx) |

    = k² e^(kx) ≠0

    故 1, e^(kx), xe^(kx)為線性獨立

    2007-12-12 21:18:36 補充:

    法三之" | " 為行列式, 右側無法對齊,請注意!

    Source(s): me
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