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#ed_op#DIV#ed_cl#又搞烏龍了！#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#希臘字w  應該是 (n-1) by (n-1)  不是 n by n。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#事實上由代數餘子式也可以看的出來會少一個order。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#實際上這個方法很技巧。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#先令特解形式為齊次解但是待定係數為x的函數  C(x)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#然後求導一次(利用Chain rule)，再令 Sum  C'(x)*y =0 (技巧)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#然後依次在求導帶入化簡就可以得到與公式相等的敘述#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#是一個  n by n 的線性方程式系統(linear equation system)#ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#無聊就寫一下VOP的公式#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#P align=left#ed_cl##ed_op#IMG alt="image file name: 2k636aaba8f2.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k636aaba8f2.png" border=0#ed_cl##ed_op#/P#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#訂正：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(5)部分那個Im跟Re寫反了，請自行修正。(很難打，馬有失蹄)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#記憶方法為靠Euler formula。#ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#沒辦法顯示那麼大請把圖抓回去再開即可(右鍵另存圖片)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此法有侷限性，僅對非齊次項為下面才可以簡便計算#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2k6e14a659bb.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k6e14a659bb.png" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#附錄：逆算子常用公式#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(5)  Re表示實部，Im表虛部#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#      另外，L下標是k的表示與(3)的表示法無異！#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#      [事實上是利用(3)算出後取實部虛部而已]       #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#       #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(6)  L(D)化為  D^2的函數，以L-hat (D^2)表示#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(Note：個人常用(5)更勝(6)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(7)L(D)表為升冪排列，最大的degree為 m#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#     因此做1除L(D)的長除法直到第m+1項出現#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#     也就是直到出現D^m項，因為D^m+1以上對多項式來說皆是0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#     考量power rule,對m次多項式求導m次為常數,對常數求導為0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(8)不常用，大多直接用VOP了！#ed_op#/DIV#ed_cl##ed_op#P align=left#ed_cl##ed_op#IMG alt="image file name: 2k29e022ce02.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k29e022ce02.png" border=0#ed_cl##ed_op#/P#ed_cl##ed_op#B#ed_cl##ed_op#FONT face=Verdana size=2#ed_cl##ed_op#P align=left#ed_cl##ed_op#IMG alt="image file name: 2k8438a677dc.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k8438a677dc.png" border=0#ed_cl##ed_op#/P#ed_cl##ed_op#P align=center#ed_cl##ed_op#B#ed_cl##ed_op#FONT face=Verdana size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/B#ed_cl# #ed_op#/P#ed_cl##ed_op#/FONT#ed_cl##ed_op#/B#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#基本上那個叫做逆算子法，只對特殊型有用。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#運用的概念是積分等於反導數，然後把相應的特殊法則記下來而已。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#解法類似這樣(其實你的東西我有一部分看不懂，不過答案都一樣)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#這種方法並不是萬用法，是要記特殊的非齊次項去做。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#P align=center#ed_cl##ed_op#IMG alt="image file name: 2k1fef4ad70c.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k1fef4ad70c.png" border=0#ed_cl##ed_op#/P#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#如題...#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#誰能把整個運算過澄清清楚的寫出來.....#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#千萬不要跳過程#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#不然看不懂 Orz.............................#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#題目：  y"' + y" = (e^x)cosx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#講義解的= =#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#  ( D^3 + D^2 ) y =(e^x)cosx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#移項成逆運算公式：#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#  Yp =  [1/ ( D^3 + D^2 )] ( e^x ) cosx #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#        = [ 1/(D+1)^3 + (D+1)^2 ] ( e^x ) cosx  ---------->為何會變這樣Orz...#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#整理#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#  Yp = e^x [ 1 / ( D^3 + 4D^2 + 5D + 2 ) ] cosx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#        =  e^x [ 1 / ( 4D - 2 ) ] cosx..........................看不懂又跳太快#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#        =  e^x [ ( 4D + 2 ) / ( 4D - 2 ) ( 4D + 2 ) ] cosx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#        =  e^x [ ( 4D + 2 ) / ( 16D - 4 ) ] cosx#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#        =  e^x [ ( 4D + 2 ) / 20 ] cosx.........................又看不懂了\ = =投降#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#        =  e^x [ ( 1 / 5 ) sinx - ( 1 / 10 ) cosx ]....................天阿根本是天書跳太快了#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#它的特解根本看不懂跳太快................#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#誰能幫我算沒跳過的全解......Orz....拜託了#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#