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# 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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- ～口卡口卡 修～Lv 51 decade agoFavorite 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

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