Re: [理工] [工數]-一階ODE
y' = (2cosx-sinx)δ(x) ; y(-1)=1
∞ -sx
sY(s)-y(0) = ∫(2cosx-sinx)δ(x)*e dx (作Laplace轉換)
0
|
= (2cosx-sinx)e^(-sx)|
|x=0
= 2
let y(0)= C
2 C
Y(s) = --- + ---
s s
y(x) = 2u(x) + C 代初值y(-1)=1
1 = 0 + C
=> C = 1
=> y(x) = 2u(x) + 1
------------------------------
Picard's Iteration
dy/dx = (2cosx-sinx)δ(x) ; y(-1)=1
dy = (2cosx-sinx)δ(x)
x x
∫ d(y(x)) = ∫ (2cosx-sinx)δ(x)dx
x0 x0
|x x
y(x)| = ∫ (2cosx-sinx)δ(x)dx ; x0 = -1
|-1 -1
x
y(x) - y(-1) = ∫ (2cosx-sinx)δ(x)dx ; y(-1)=1
-1
x
y(x) = 1 + ∫ (2cosx-sinx)δ(x)dx
-1
討論
(1)x<0
y(x) = 1 + 0 = 1
(2)x=0
y(x) = 1 + 1 = 2
(3)x>0
y(x) = 1 + 2 = 3
總結(1)(2)(3)
y(x) = 1 + 2u(x)
--
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※ 編輯: HerbyEli 來自: 118.169.135.197 (01/10 12:40)
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