可以幫我解一下這題嗎?
(x +y)^2 y"-2(x+ 1)y'+ 2y=0,其中一解為y=x +1
謝謝~~
2005-12-30 15:52:40 · 1 個解答 · 發問者 Anonymous in 教育與參考 ➔ 考試
阿伯版主,這題給一個已知的解,是〝解〞,不是〝特解〞喔!所以用 reduction of order來做。在考場上我們就不要去推導方法的來由,所以,以下的 reduction of order方法是我們要記的! 一高階ODE,型式為:y'' + P(x)y' + Q(x)y = 0 且給定一已知解y1求另一解y2,求法:U = [ 1/( y1 )2 ]e∫- Pdxy2 = y1∫Udx1. Solve ( x + 1)2y" - 2( x + 1 )y' + 2y = 0 has y1= x + 1Sol: 原式化為:y'' - [ 2/( x + 1 ) ]y' + [ 2/( x + 1 )2 ]y = 0 U = [ 1/( y1 )2 ]e∫- Pdx = [ 1/( x + 1 )2 ]e∫2/( x + 1 )dx = [ 1/( x + 1 )2 ]e 2‧ln│x + 1│ = [ 1/( x + 1 )2 ][ (x + 1 )2 ] = 1 y2 = y1∫Udx = ( x + 1 )∫1‧dx = x( x + 1 ) y = y1 + y2 → y = x + 1 + x( x + 1 ) = ( x + 1 )( x + 1 ) = ( x + 1 )2 → y = ( x + 1 )2 # 解完了,有點在記憶公式,但考場上實在沒太多時間推導,以上方法才正確,不要用什麼〝因變數變換〞去做,一定做不出來,在考場上選錯解題方法,付出的代價都很大,要小心喔!
2005-12-31 20:16:52 · answer #1 · answered by 龍昊 7 · 0⤊ 0⤋