Re: [其他] 矩陣+ode
※ 引述《gggg9999 (居九)》之銘言：
: 解下列方程式
: 4 2 3e^(t)
: X'=[ ]X+[ ]
: 2 1 e^(t)
: 另外請問 無限大 dx
: [ ---------
: 0 1+x^(4)
: 如何解呢 是要用復利葉嗎@@
: 感謝大大了!!
1.
4 2 T
Let x'=Ax+b , A=[ ] = SDS
2 1
其中
1 1 2 0 0 3e^t
S=----[ ] , D=[ ],b=[ ]
√5 -2 1 0 5 e^t
Let x=Sy
帶入原式
T
Sy'=SDS Sy+b
=> T
y'=Dy+S b
=>
y1'=(1/√5)*e^t
y2'=5y2 + (7/√5)*e^t
=>
y1=(1/√5)*e^t+c1
(D-5)y2=(7/√5)*e^t , D表對t微分運算子
=>
y1=c1+(1/√5)*e^t ...(1)
y2=c2*e^5t-(7/4√5)*e^t ...(2)
又 x=Sy
所以
x1=(1/√5)y1+(2/√5)y2
x2=(-2/√5)y1+(1/√5)y2
代入(1).(2)即為解
∞ 1 ∞ 1
2.∫ -------- dx = 0.5∫ ---------dx = 0.5*K
0 x^4 + 1 -∞ x^4 + 1
利用留數定理
1
K + ∫---------dz =2πi[Res(e^iπ/4)+ Res(e^i3π/4)]
c z^4 + 1
ps.極點為z^4=e^i(2n+1)π
=>z=e^i((2n+1)/4)π,n=0.1.2.3 四個極點
其中,上半圓內極點為n=0.1,取Res(e^iπ/4).Res(e^i3π/4)
由於 deg(x^4+1)= 4 > deg(1)+1 = 1
1
所以∫---------dz = 0
c z^4 + 1
故
K=2πi[Res(e^iπ/4)+ Res(e^i3π/4)]
=2πi[lim (z-e^iπ/4)*(1/(z^4+1)) + lim (z-e^iπ/4)*(1/(z^4+1))]
z->e^iπ/4 z->e^i3π/4
=2πi[lim 1/(4z^3) + lim 1/(4z^3) ]
z->e^iπ/4 z->e^i3π/4
=(π/2)i[exp(-i3π/4) + exp(-i9π/4)]
=(π/2)i[-i/√2 -i/√2]
=π/√2
原式=0.5K=π/(2√2)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.113.123.237
推
02/29 04:23, , 1F
02/29 04:23, 1F
推
02/29 08:20, , 2F
02/29 08:20, 2F
→
02/29 08:39, , 3F
02/29 08:39, 3F
→
02/29 08:39, , 4F
02/29 08:39, 4F
→
02/29 08:40, , 5F
02/29 08:40, 5F
→
02/29 12:23, , 6F
02/29 12:23, 6F
→
02/29 12:23, , 7F
02/29 12:23, 7F
→
02/29 12:24, , 8F
02/29 12:24, 8F
→
02/29 12:25, , 9F
02/29 12:25, 9F
→
02/29 12:25, , 10F
02/29 12:25, 10F
推
02/29 20:37, , 11F
02/29 20:37, 11F
討論串 (同標題文章)