推 yuxike:這解法好特別0.0 沒做過想不出來耶 感謝大大^^ 09/11 02:11
※ 引述《yuxike (非人Zai)》之銘言：
: 小弟又來了
: 每次都是棘手的題目(其實是花時間的題目)
: X1"(t) = (-13/2)X1 + (5/2)X2 + 2[1-U(t-3)]
: X2"(t) = (5/2)X1 + (-13/2)X2
: X1(0) = X2(0) = X1'(0) = X2'(0) = 0
: ANS:
: X1(t)= (13/36) + (-1/4)cos2t + (-1/9)cos3t
: -[(13/36) + (-1/4)cos2(t-3) + (-1/9)cos3(t-3)]U(t-3)
: X2(t)= (5/36) + (-1/4)cos2t + (1/9)cos3t
: -[(5/36) + (-1/4)cos2(t-3) + (1/9)cos3(t-3)]U(t-3)
: 我可以用wronskian算出X2(s)...
: 再逆轉換可得X2(t)與答案相符
: 但用X2(t)帶回原式，好像沒辦法藉消去法求出X1(t)
: 而答案的X1(t)轉LAPLACE之後
: 跟我wronskian算出來的X1(s)不同...
: 所以上來問看看，用什麼方法可以解出來
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我把 para. ( X1(t) , X2(t) ) 寫成 ( x(t) , y(t) )
單純不想寫下標而已 XD
┌ x'' = (-13/2)x + (5/2)y + 2[1-u(t-3)] ____(1)
└ y'' = (5/2)x + (-13/2)y ____(2)
(1) + (2) :
(x+y)'' = -4(x+y) + 2[1-u(t-3)] with (x+y)(0) = (x+y)'(0) = 0
可解得
x+y = m[1 - cos(2t)] - m{1 - cos[2(t-3)]}*u(t-3) ____(3)
where m = (1/2)
相同的， (1) - (2):
(x-y)'' = -9(x-y) + 2[1-u(t-3)] with (x-y)(0) = (x-y)'(0) = 0
可解得
x-y = n[1 - cos(3t)] - n{1 - cos[3(t-3)]}*u(t-3) ____(4)
where n = (2/9)
因此由 (3)(4) 可解出:
┌ x(t) = (m+n)/2 - (m/2)cos(2t) - (n/2)cos(3t)
│ -(1/2){ (m+n) - m*cos[2(t-3)] - n*cos[3(t-3)] }*u(t-3)
│
└ y(t) = (m-n)/2 - (m/2)cos(2t) + (n/2)cos(3t)
-(1/2){ (m-n) - m*cos[2(t-3)] + n*cos[3(t-3)] }*u(t-3)
把 (m,n) = (1/2 , 2/9) 帶入就是答案了
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