Re: [微積] 2階ODE-未定係數解(特殊題目)

看板Math作者Heaviside (Oliver)時間12年前 (2013/09/28 15:41)推噓2(2推 0噓 12→)

留言14則, 7人參與討論串2/2 (看更多)

※ 引述《pigheadthree (爬山)》之銘言： : 題目：y""-3y'+2y = 8x^2 - 2x^(e^x) : 這是小弟的筆記上的題目，沒有答案，以下為小弟的解法， : 不知道計算過程是否有問題，麻煩版上前輩們不吝嗇指導，謝謝！原文43 話說到底是四階ODE 還是二階ODE呀..... ---- y"-3y'+2y=8x^2-2x^(e^x) 由觀察法得y_h=C1exp(2x)+C2exp(x) 令y_p=ψ1(x)y1(x)+ψ2(x)y2(x) y_p'=ψ1(x)y'1(x)+ψ2(x)y'2(x)+ψ'1(x)y1(x)+ψ'2(x)y2(x) 令ψ'1(x)y1(x)+ψ'2(x)y2(x)=0 y_p"=ψ1(x)y"1(x)+ψ2(x)y"2(x)+ψ'1(x)y'1(x)+ψ'2(x)y'2(x) 將y_p、y_p'、y_p"代回得 [ψ1(x)y"1(x)+ψ2(x)y"2(x)+ψ'1(x)y'1(x)+ψ'2(x)y'2(x)] -3[ψ1(x)y'1(x)+ψ2(x)y'2(x)]+2[ψ1(x)y1(x)+ψ2(x)y2(x)]=8x^2-2x^(e^x) 超煩另解使用Heaviside求解法 d let D= ── dx 原式=>(D-2)(D-1)y= 8x^2 - 2x^(e^x) y_h=C1exp(2x)+C2exp(x) 1 y_p= ───── [ 8x^2 - 2x^(e^x)]=y_p1-y_p2 (D-2)(D-1) 1 y_p1= ─────(8x^2)=(0.5+0.75D+0.875D^2+...)(8x^2) D^2-3D+2 =4x^2 +6x+7 1 1 1 y_p2 = ───── [2x^(e^x)]=[─── - ───][2x^(e^x)] (D-2)(D-1) D-2 D-1 =exp(2x)∫exp(-2x)[2x^(e^x)]dx - exp(x)∫exp(-x)][2x^(e^x)]dx 因y_p2無法積分丟在那裡就好得解y=y_h+y_p ----- 若題目改為 y"-3y'+2y = 8x^2 - 2x(e^x) 則y_p2 =exp(2x)∫exp(-2x)[2x*(e^x)]dx - exp(x)∫exp(-x)][2x*(e^x)]dx =exp(x)[-x^2-2x-1] 得解y=y_h+y_p = C1exp(2x)+C2exp(x)+4x^2 +6x+7+exp(x)[x^2+2x]為解強烈懷疑原PO抄錯題目了@@ -- Logic can be patient for it is eternal. ----- Oliver Heaviside -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.185.134.5

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LPH66

09/28 15:43, , 1^F

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Heaviside

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09/28 20:55, , 3^F

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