推 LaAc : 謝謝你！ 11/02 08:43
※ 編輯: Vulpix (61.230.70.164), 11/11/2018 01:09:50
※ 引述《LaAc (深淵戰士)》之銘言：
: http://i.imgur.com/6DGdYwz.jpg
: http://i.imgur.com/h7AwdW2.jpg
: 想請教這題的特解要怎麼求。
: （第二張圖是我求通解的算式。）
: 我試過用A（x）e^x代入求，但沒有用，最後會開不了。
: 請問還有沒有什麼方法可以求？
: 我問過好幾個人都說不會這種形式的QQ
1. variation of constant
Let y = A(x)+B(x)x+C(x)/x
=> y' = A'+B'x+C'/x + B-C/x^2 = B-C/x^2 (assuming that A'+B'x+C'/x=0)
=> y" = B'-C'/x^2 + 2C/x^3 = 2C/x^3 (assuming that B'-C'/x^2=0)
=> y"'= 2C'/x^3 - 6C/x^4
And xy"'+3y"=e^x becomes 2C'/x^2=e^x
=> C'=0.5x^2e^x => B'=C'/x^2=0.5e^x => A'=-xe^x
So C=(0.5x^2-x+1)e^x+c, B=0.5e^x+b, A=(-x+1)e^x+a
=> y = A(x)+B(x)x+C(x)/x = e^x/x + a + bx + c/x
2. integrating factor
(x^3y")' = x^3y"'+3x^2y" = x^2e^x
=> x^3y" = (x^2-2x+2)e^x+c
=> y" = (1/x-2/x^2+2/x^3)e^x+c/x^3
=> y' = ∫(1/x-2/x^2+2/x^3)e^xdx - 0.5c/x^2
= ∫(e^x/x)dx + ∫(-2/x^2+2/x^3)e^xdx - 0.5c/x^2
= ∫(e^x/x)dx + (2/x-1/x^2)e^x - ∫(2/x-1/x^2)e^xdx - 0.5c/x^2
= ∫(-e^x/x)dx + (2/x-1/x^2)e^x + ∫(1/x^2)e^xdx - 0.5c/x^2
= ∫(-e^x/x)dx+(2/x-1/x^2)e^x+(-1/x)e^x-∫(-1/x)e^xdx -0.5c/x^2
= (1/x-1/x^2)e^x -0.5c/x^2 + b
=> y = ∫(1/x-1/x^2)e^xdx + 0.5c/x + bx
= ∫(e^x/x)dx - ∫(e^x/x^2)dx + 0.5c/x + bx
= ∫(e^x/x)dx - (-e^x/x) +∫(-e^x/x)dx + 0.5c/x + bx
= e^x/x + 0.5c/x + bx + a
3. series solution
Solve for xy_n"'+3y_n" = x^n => (x^3y_n")' = x^{n+2}
=> x^3y_n" = x^{n+3}/{n+3} + c_n
=> y_n" = x^n/{n+3} + c_n/x^3
=> y_n = x^{n+2}/{(n+1)(n+2)(n+3)} + 0.5c_n/x + b_nx + a_n
Then y = Σ_{n=0}^{∞} {y_n/n!}
= Σ_{n=0}^{∞} {x^{n+2}/(n+3)! + 0.5c_n/x + b_nx + a_n}
= (e^x-1-x-0.5x^2)/x + 0.5Σc_n/x + Σb_{n}x + Σa_n
= e^x/x + (-1+0.5Σc_n)/x + (-0.5+Σb_n)x -1+Σa_n
All these solutions coincide, as we expected.
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