求解二階尤拉科西方程

(1) 齊次解令 y = x^m則 y' = m*x^(m-1), y" = m(m-1)*x^(m-2)代入原式裡面:0 = m(m-1) - 5m + 8= m^2 - 6m + 8= (m - 2)(m - 4)m = 2, 4
yh(x) = c1*x^2 + c2*x^4
(2) 特殊解yp = a*ln(x) + bx + cyp' = a/x + byp" = -a/x^2 代入原式裡面:2*ln(x) = -a - 5a - 5bx + 8a*ln(x) + 8bx + 8c= (-6a + 8c) + 3bx + 8a*ln(x)8a = 2 ==> a = 1/43b = 0 ==> b = 0c = 3a/4 = 3/16==> yp(x) = 0.25*ln(x) + 3/16
(3) 一般解y(x) = c1*x^2 + c2*x^4 + 0.25*ln(x) + 3/16
(3.a) 初始條件:y(0) = 0 + 0 + 0.25*ln(0) + 3/16 = 3/16==> 0.25*ln(0) = 0 = 錯誤
(3.b) 邊界條件:y(e) = c1*e^2 + c2*e^4 + 0.25*ln(e) + 3/16= c1*e^2 + c2*e^4 + 1/4 + 3/16= c1*e^2 + c2*e^4 + 7/16= e^2 - e^4 + 7/16 ==> c1 = 1, c2 = -1==> y(x) = x^2 - x^4 + 0.25*ln(x) + 3/16
(4) 建議取消 y(0) = 3/16

2015-07-06 16:10:12 補充：
改為 y(1) = 3/16 ==> c1 + c2 = 0

初始條件 有誤喔 大大

https://www.wolframalpha.com/input/?i=%28x%5E2%29y%27%27-5xy%27%2B8y%3D2ln%28x%29

2015-07-05 23:36:47 補充：
方程有 2ln(x)，可以代入 x = 0 嗎？？？

2015-07-05 23:39:09 補充：
若你當 ln(0) = 0 的話，那麼解答是：

y = x² - x⁴ + ln(x)/4 + 3/16

2015-07-05 23:39:24 補充：
https://www.wolframalpha.com/input/?i=simplify+%28x%5E2%29%28x%5E2+-+x%5E4+%2B+ln%28x%29%2F4+%2B+3%2F16%29%27%27-5x%28x%5E2+-+x%5E4+%2B+ln%28x%29%2F4+%2B+3%2F16%29%27%2B8%28x%5E2+-+x%5E4+%2B+ln%28x%29%2F4+%2B+3%2F16%29

2015-07-07 00:16:24 補充：
對，麻辣長英明，y(1) = 3/16 即可！