1.Find the first solution xy"+2y'-xy=0 [20point]
2.From the generating function of the Legendre polynomial
1 ∞
g(x,t)=------------- Σ Pn(x)t^n ,
√(1-2xt+t^2) n=0
show that (1) P(0)=(-1)^n．1．3．5…(2n-1)/(2．4．6…2n)
2n
1 2
(2) ∫P^2(x)dx = ----- [30point]
-1 n 2n+1
3.Solve y"-xy'-y=0 [20point]
Note:
1 1 1
ln(1-x)=x - ---(x^2) - ---(x^3) - ---(x^4) - …
2 3 4
1
-----=Σx^n
1-x
1 1 1．3 1．3．5 1．3．5．7
---------- = 1 + ---x + ----(x^2) + -------(x^3) + ---------- + …
(1-X)^(1/2) 2 2．4 2．4．6 2．4．6．8
＊ 2.(1)(2)的2n及n為P的下標
po一篇文章花了快1小時.....
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※ 編輯: fgfdsa 來自: 59.117.126.150 (01/14 02:35)
※ 編輯: fgfdsa 來自: 59.117.126.150 (01/14 02:36)