[微積] 2階ODE-未定係數解(特殊題目)
題目:y"-3y'+2y = 8x^2 - 2x^(e^x)
這是小弟的筆記上的題目,沒有答案,以下為小弟的解法,
不知道計算過程是否有問題,麻煩版上前輩們不吝嗇指導,謝謝!
設 y = ax^2 + bx + c
y'= 2ax + b
y''= 2a
(2a)x^2 -3*(2a+b)x + (ax^2 + bx +c) = 8x^2 - 2x^(e^x)
2ax^2 - 6ax^2 - 3bx + ax^2 + bx + c - 8x^2 + 2x^(e^x) = 0
(-3a-8)x^2 - 2bx + c + 2x^(e^x) = 0
a = - (8/3)
b = 0
c = 0
-(8/3)x^2 + 2x^(e^x) = 0
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常係數計算:
設 x^2 - 3x + 2 = 0
(x-1)*(x-2) = 0
x = 1 or 2
常係數:φ1 = e^x 與 φ2 = e^2x
所以方程解為:y = -(8/3)x^2 + 2x^(e^x) + c1*e^x + c2*e^2x (通解)
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這是小弟計算最後的答案,不知道是否有問題?
麻煩不吝嗇指導,謝謝!
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◆ From: 114.35.30.78
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不好意思,可以請教您的計算方式嗎?
y = c1*e^x + c2*e^2x + 4x^2 + 12x + 14
+ 2e^x[∫(x^e^x)*(e^-x) dx] - 2e^2x[∫(x^e^x)*(e^-2x) dx]
以上式子怎麼出來的?
※ 編輯: pigheadthree 來自: 114.35.30.78 (09/27 21:21)
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不好意思,題目打錯了,為2階ODE,已作修正。
※ 編輯: pigheadthree 來自: 114.35.30.78 (09/28 19:38)
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7年前
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6年前
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