Derek asked in 科學數學 · 1 decade ago

求二階變係數ODE?

求(1− 2x) y"+ 2 y'+ (2x − 3)y = 0

答:y = ce^x + kxe^− x

想請教這題該怎麼解?!

2 Answers

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  • 1 decade ago
    Favorite Answer

    (1− 2x) y"+ 2 y'+ (2x − 3)y = 0

    → (1− 2x)( y" - y') + (1− 2x) y' + 2 y'+ (2x − 3)y = 0

    → (1 - 2x)( y" - y') + (3 - 2x)( y' - y) = 0

    令 k(x) = y' – y ,則可將上式改寫成:

    (1 - 2x)*k' + (3 - 2x)*k = 0

    → 1/k dk = (2x-3)/(1-2x) dx

    → ∫1/k dk = ∫- 1 - 2/(1-2x) dx

    → lnk = - x + ln(1-2x) + C1

    → k = C*(1-2x)*exp(- x) ____where C1 and C are integrating constant

    → y' – y = C*(1-2x)*exp(- x)

    → [y*exp(-x)] ' = C*(1-2x)*exp(- 2x)

    → y*exp(-x) = ∫C*(1-2x)*exp(- 2x) dx

    = C*x*exp(-2x) + C2

    → y = C*x*exp(- x) + C2*exp(x)

    2008-05-25 16:31:37 補充:

    忘了打

    where C2 is a integrating constant

  • 1 decade ago

    阿祥大大 看了說明 又多學了一招變係數的解法了

    感謝了.....

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