工程數學禮拜二要考,緊急求救,一題20點,有5題喔
x^3y''' xy' - y = x^2
y(1) = 0 y'(1) = 3 y''(1) = 3
2006-01-12 18:08:56 · 2 個解答 · 發問者 Anonymous in 科學 ➔ 工程學
忘記補加號
x^3y'''+xy'-y=x^2
2006-01-12 18:26:43 · update #1
5. Solve x3y''' + xy' - y = x2 , y(1) = 0 , y'(1) = 3 , y''(1) = 3sol: 令 x = et → t = ln│x│ y' = dy/dx = ( dy/dt )( dt/dx ) = ( 1/x )( dy/dt ) y'' = d2y/dx2 = ( d/dx )[ ( 1/x )( dy/dt ) ] = ( 1/x2 )( d2y/dt2 ) - ( 1/x2 )( dy/dt ) y''' = d3y/dx3 = ( d/dx )[ ( 1/x2 )( d2y/dt2 ) - ( 1/x2 )( dy/dt ) ] = ( 1/x3 )( d3y/dt3 ) - ( 3/x3 )( d2y/dt2 ) + ( 2/x3 )( dy/dt ) 將 y'、y'''、x = et 代入原 Cauchy - Euler equation 得: → ( d3y/dt3 ) - 3( d2y/dt2 ) + 3( dy/dt ) - y = e2t 特徵方程式:r3 - 3r2 + 3r - 1 = 0 → ( r - 1 )( r - 1 )( r - 1 ) = 0 → r = 1 , 1 , 1 ~ 三重根 yh = ( c1 + c2t + c3t2 )et = ( c1 + c2ln│x│+ c3ln2│x│)x ~ 齊性解 用未定係數法求特解: 令 yp = Ae2t → dyp/dt = 2Ae2t , d2yp/dt2 = 4Ae2t , d3yp/dt3 = 8Ae2t ( d3yp/dt3 ) - 3( d2yp/dt2 ) + 3( dyp/dt ) - yp = e2t → Ae2t = e2t 比較係數得:A = 1 → yp = e2t = x2 ~ 特解 通解:y = yh + yp → y = ( c1 + c2ln│x│+ c3ln2│x│)x + x2 代入邊界條件 x = 1 , y = 0 得: c1 + 1 = 0 → c1 = - 1 y' = [ ( c2/x ) + 2c3ln│x│/x ]x + ( c1 + c2ln│x│+ c3ln2│x│) + 2x 代入邊界條件 x = 1 , y = 3 得: c2 + c1 + 2 = 3 → c2 = 3 - 2 - c1 = 2 y'' = [ - c2/x2 + ( 2c3/x - 2c3ln│x│)/x2 ]x + 2( c2/x + 2c3ln│x│) + 2 代入邊界條件 x = 1 , y = 3 得: - c2 + 2c3 + 2c2 + 2 = 3 → c3 = ( 1 - c2 )/2 = - 1/2 → y = [ - 1 + 2ln│x│- ( 1/2 )ln2│x│]x + x2 #
2006-01-13 16:45:00 補充:
匿名版主您好,其實這些題目都可以技巧性避開以參數變異法求特解,這樣計算上就可以少了很多積分步驟,比較簡單,你們老師勾這五題都很能測出你們的程度,好好加油!
2006-01-12 19:24:18 · answer #1 · answered by 龍昊 7 · 0⤊ 0⤋
神~你好神~
2006-01-15 03:28:37 · answer #2 · answered by 流光 3 · 0⤊ 0⤋