Research Space for Linear Algebra & Discrete Mathematics
x2' = -4x2 - 4e^(-2t)特微方程式a + 4 = 0=> a = -4齊次解x2(h) = ce^(-4t)令特解x2(p) = d e^(-2t)代入原式-2d e^(-2t) = -4(d e^(-2t)) - 4e^(-2t)=> -2d = -4d - 4=> d = -2=> x2(p) = -2 e^(-2t)則x2 = x2(h) + x2(p) = c e^(-4t) - 2e^(-2t)
x1' = -2x1 - e^(-2t)特徵方程式a + 2 = 0=> a = -2=> x1(h) = c e^(-2t)令x1(p) = d t e^(-2t)代入原式d e^(-2t) - 2d t e^(-2t)= -2d t e^(-2t) - 2e^(-2t)=> d - 2dt = -2dt - 2=> d = -2=> x1(p) = -2te^(-2t)=> x1 = x1(h) + x1(p) = c e^(-2t) - 2te^(-2t)這些觀念會用到微分方程
哦~我會了謝謝老師
x2' = -4x2 - 4e^(-2t)
回覆刪除特微方程式a + 4 = 0
=> a = -4
齊次解x2(h) = ce^(-4t)
令特解x2(p) = d e^(-2t)代入原式
-2d e^(-2t)
= -4(d e^(-2t)) - 4e^(-2t)
=> -2d = -4d - 4
=> d = -2
=> x2(p) = -2 e^(-2t)
則x2 = x2(h) + x2(p)
= c e^(-4t) - 2e^(-2t)
x1' = -2x1 - e^(-2t)
回覆刪除特徵方程式a + 2 = 0
=> a = -2
=> x1(h) = c e^(-2t)
令x1(p) = d t e^(-2t)代入原式
d e^(-2t) - 2d t e^(-2t)
= -2d t e^(-2t) - 2e^(-2t)
=> d - 2dt = -2dt - 2
=> d = -2
=> x1(p) = -2te^(-2t)
=> x1 = x1(h) + x1(p)
= c e^(-2t) - 2te^(-2t)
這些觀念會用到微分方程
哦~我會了
回覆刪除謝謝老師