推 cccoco:謝謝d大@@" 03/09 10:14
※ 引述《cccoco (危機感)》之銘言：
: 請問
: x'=3x-y-z
: y'=x+y-z+t
: z=x-y+z+2e^t
: 該怎麼解比較快呀..
: 我用克拉瑪解超久的...
: 謝謝喔@@
: 另外請問一下
: 當取col(A)或row(A) 是要取化檢後的那行還是化簡的呢@@?
: 我看周易兩種都取過
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┌ x' = 3x - y - z ____(1)
│ y' = x + y - z + t ____(2)
└ z' = x - y + z + 2e^t ____(3)
由 (1)-(2) → ┌ (x-y)' = 2(x-y) - t
(1)-(3) └ (x-z)' = 2(x-z) - 2e^t
→ ┌ x - y = c1*e^(2t) + (2t+1)/4 ____(4)
└ x - z = c2*e^(2t) + 2e^t ____(5)
把 (4)(5) 帶入 (1):
x' = x + (c1+c2)*e^(2t) + (2t+1)/4 + 2e^t
→ x = (c1+c2)*e^(2t) - (2t + 3)/4 + 2te^t + c3*e^t ____(6)
再將 (6) 分別帶回 (4)(5) 解得:
┌ y = c2*e^(2t) - (t + 1) + 2te^t + c3*e^t ____(7)
└ z = c1*e^(2t) - (2t + 3)/4 + 2te^t + (c3-2)*e^t ____(8)
(6)(7)(8) 即為通解
ps:
或寫成 ┌ x ┐ ┌ 1 ┐ ┌ 1 ┐ 2t ┌ 1 ┐ t
│ y │ = (c1│ 0 │ + c2│ 1 │)e + c3│ 1 │e
└ z ┘ └ 1 ┘ └ 0 ┘ └ 1 ┘
┌ 2te^t - (2t + 3)/4 ┐
+ │ 2te^t - (t + 1) │
└ 2(t-1)e^t - (2t + 3)/4 ┘
比較好看 XDD
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