Re: [理工] [工數] [ode]

看板Grad-ProbAsk作者doom8199 (～口卡口卡　修～)時間13年前 (2011/01/06 16:02)推噓5(5推 0噓 0→)

留言5則, 5人參與討論串4/4 (看更多)

※ 引述《asdf322505 ()》之銘言： : 2x 2 : y''— ------y'+ ----- y =12(x^2+1) : x^2+1 x^2+1 : ans:y=c1x+c2(x^2-1)+2x^4+6x^2 : 這題 : 想好久 : 誰能幫我解答@@ --- 1 令 m = ──── x^2 + 1 所以原 ode → y'' - 2xmy' + 2my = 12/m → (xy'-y)'/x - 2m(xy'-y) = 12/m → m(xy'-y)' - 2x(m^2)(xy'-y) = 12x → m(xy'-y)' + m'(xy'-y) = 12*x since m' = -2x(m^2) → [m(xy'-y)]' = 12x → [m(x^2)(y/x)']' = 12x → m(x^2)(y/x)' = 6x^2 + c1 → (y/x)' = 6(x^2 + 1) + c1*[1 + 1/(x^2)] → y = (2x^4 + 6x^2) + c1*(x^2 - 1) + c2*x ps: 也能用 [(x-i)D - 1][(x+i)D - 2]y = 12(x^2+1)^2 (x+i)^3 y → { ────*[────]' }' = 12(x+i)^2 (x-i) (x+i)^2 (x+i)^3 y → ────*[────]' = 4(x+i)^3 + c1 (x-i) (x+i)^2 y 1 -2i → [────]' = 4(x-i) + c1*[ ──── + ──── ] (x+i)^2 (x+i)^2 (x+i)^3 y -1 i → ──── = 2(x-i)^2 + c1*[ ─── + ──── ] + c2 (x+i)^2 (x+i) (x+i)^2 → y = 2(x^2 + 1)^2 + c1*(-x) + c2*(x+i)^2 = 2(x^2 + 1)^2 + c1'*x + c2'*(x^2 - 1) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.211.136

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