推 t5d:Thanks~ 03/05 07:35
※ 引述《t5d (t5d)》之銘言：
: 1.題目要求用拉氏解
: x' - 2y' + 3z = 0
: x - 4y' + 3z'= t
: x - 2y' + 3z'= -1
: x(0) = y(0) = z(0) = 0
: 囧> 只會用2.3式相減求出y 在代回去就不知道怎麼求x z了
: 2.Evaluate the function by using the Residue theorem
: 2π cos2θ
: ∫ ------------------ dθ , p > 1
: 0 1 - 2p cosθ + p^2
令 p ≠ 1 且 z = e^iθ , dθ = dz/iz 帶回
2π 1/2( z^2 + 1/z^2 ) dz
I = ∫ --------------------------------- ----
0 (1 + p^2 ) - 2p(1/2)( z + 1/z ) iz
1 2π ( z^4 + 1 )/z^2
= ----∫ --------------------------- dz
2i 0 (1 + p^2 ) - p( z^2 + 1 )
-1 2π z^4 + 1
= -----∫ ------------------------------- dz
2pi 0 z^2[ z^2 - ( 1/p + p )z + 1 ]
i 2π z^4 + 1
= ----∫ ------------------------- dz
2p 0 z^2( z - 1/p )( z - p )
z^4 + 1
令 f(z) = -------------------------
z^2( z - 1/p )( z - p )
在 C : |z| = 1內具有 z = 0的2階pole 及 z = 1/p的1階pole ( p > 1 )
d 1 + p^2
Res f(0) = lim ----[ z^2f(z) ] = ---------
z→0 dz p
1 1 1 + p^4
Res f(---) = lim ( z - --- )f(z) = --------------
p z→1/p p p( 1 - p^2 )
i 1
所求 I = ---- 2πi[ Res f(0) + Res f(---) ]
2p p
-2π
= ----------------
p^2( 1 - p^2 )
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