Re: [理工] [工數]-Laplace 解聯立ODE

看板Grad-ProbAsk作者ytyty (該換個版潛水了™ )時間14年前 (2010/03/28 01:07)推噓2(2推 0噓 4→)

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※ 引述《kissboy1065 ( 川)》之銘言： : dx/dt + x + y = e^-3t ,x(0)=1 : dy/dt + 4x +y = 0 ,y(0)=0 : 麻煩你了> < 兩邊取Laplace transform得 1 (s+1)X - x(0) + Y = ── s+3 4X + (s+1)Y - y(0) = 0 整理得 s+4 (s+1)X + Y = ── s+3 4X + (s+1)Y = 0 │s+1 1 │ Δ=│ │=(s+1)^2 -4 = s^2+2s-3 = (s+3)(s-1) │4 s+1│ s+4 │─── 1│ (s+4)(s+1) s^2+5s+4 Δ =│ s+3 │= ────── = ────── X │ │ s+3 s+3 │ 0 s+1│ s+4 │ s+1 ─── │ 4(s+4) 4s+16 Δ =│ s+3 │= ─── = ─── Y │ │ s+3 s+3 │ 4 0 │ s^2+5s+4 A B C X = ─────── = ── + ─── + ── (s+3)^2*(s-1) s+3 (s+3)^2 s-1 去分母得 s^2+5s+4 = A(s+3)(s-1) + B(s-1) + C(s+3)^2 = A(s^2+2s-3)+Bs-B+C(s^2+6s+9) = (A+C)s^2 + (2A+B+6C)s + (-3A-B+9C) 比較係數得 A+C=1 A=3/8 2A+B+6C=5 解得 B=1/2 -3A-B+9C=4 C=5/8 s^2+5s+4 3/8 1/2 5/8 X = ─────── = ── + ─── + ── (s+3)^2*(s-1) s+3 (s+3)^2 s-1 -1 -3t -3t t x(t)=L [X(s)]= (3/8)e + (1/2)te + (5/8)e 4s+16 D E F Y = ─────── = ── + ─── + ── (s+3)^2*(s-1) s+3 (s+3)^2 s-1 去分母得 4s+16 = D(s+3)(s-1) + E(s-1) + F(s+3)^2 = D(s^2+2s-3)+Es-E+F(s^2+6s+9) = (D+F)s^2 + (2D+E+6F)s + (-3D-E+9F) 比較係數得 D+F=0 D=-5/4 2D+E+6F=4 解得 E=-1 -3D-E+9F=16 F= 5/4 4s+16 -5/4 -1 5/4 Y = ─────── = ── + ─── + ── (s+3)^2*(s-1) s+3 (s+3)^2 s-1 -1 -3t -3t t y(t)=L [Y(s)]= (-5/4)e -te + (5/4)e -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 125.231.96.202

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ntust661

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ytyty

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