推 profyang :我沒記錯的話是貝索 要用power series解 12/13 18:56
推 Frobenius :要用 Frobenius Method 去解可得 BesselJn,n= ±1/2 12/13 19:38
推 Frobenius :此時α=0,β=1,λ=1,μ= ±(1/2)=> BesselJ[±1/2,x] 12/13 19:56
→ Frobenius :指標方程：r(r-1) + r - 1/4 = 0 => r= ±(1/2) 12/13 19:58
→ Frobenius :此時可以偷吃步：Let y(x) = x^(-1/2)z(x) 12/13 19:59
→ Frobenius :z(x)滿足z''(x)+z(x)=0 => z(x) = C1 cosx + C2 sinx 12/13 20:01
→ Frobenius :yh(x) = x^(-1/2)(C1 cosx + C2 sinx) 12/13 20:01
推 Frobenius :Let yp(x) = x^(1/2)(A cosx + B sinx) 代入原式 12/13 20:06
→ Frobenius :得到 2x^(3/2)(B cosx - A sinx) => A = 0 , B = 1/2 12/13 20:07
推 Frobenius :y = x^(-1/2)[ C1 cosx + (C2 + x/2 sinx) ] 12/13 20:10
推 Heaviside :Frobenius用Frobenius解Frobenius 奇哉奇哉 12/13 20:16
推 whalelover :樓上Heaviside Forbenius幫我解微方~XDD 12/13 20:32
→ Heaviside :有唷@@ 什麼時候?? 12/13 20:37
推 jack750822 :他是說兩個大師來幫他 哈哈哈 12/13 20:38
推 Heaviside :我也只會微方@@" 12/13 20:41
→ nnn820328 :謝謝幫忙,不過這題出現在參數變異法的習題中 卻要用 12/13 21:34
→ nnn820328 :Frobenius這個方式卻是沒交過的,開始來慢慢的來理解 12/13 21:36
→ nnn820328 :這過程 謝謝 12/13 21:36
推 Frobenius :XD 參數變異法主要是求解yp的另一方法，結果還是一樣 12/13 21:39
推 Frobenius :Heaviside用Heaviside解Heaviside 期待期待 12/13 21:41
→ nnn820328 :恩 不過齊次解都解不出來了 yp因該就出不來了吧? 12/13 21:41
推 Frobenius :因變數變換法 (D'Alembert 降階法) 12/13 21:44
推 Frobenius :P = x/x^2 = 1/x ; Q = (x^2 - 1/4)/x^2 12/13 21:52
→ Frobenius :u = Exp[∫(-P/2)dx] = 1/√x 12/13 21:55
推 Frobenius :R =x^(3/2)cosx/x^2,R/u=x^(3/2)cosx/x^(3/2) = cosx 12/13 22:00
→ Frobenius :C = Q - P^2/4 - P'/2 = 1 => v'' + v = R/u = cosx 12/13 22:01
推 Frobenius :v = C1 cosx + C2 sinx + x/2 sinx => y = uv 12/13 22:03
→ nnn820328 :降階法就容易懂得多了 差不多會下手了 謝謝 12/13 22:16
→ nnn820328 :不過以上方法我們老師都沒講過 就跳去講拉普拉斯轉 12/13 22:19
→ nnn820328 :換了 12/13 22:20
推 Heaviside :我已經用過好幾次Heaviside cover-up method啦XD 12/14 00:55