2
x y" + xy' + 9λy = 0
t
Let x = e , t = lnx (x>0)
d dt d -1 d
--- = --- --- = x D ( D為 --- )
dx dx dt dt
2
d d -1 -2 -2 2 -2
--- = --- (x D) = - x D + x D = x D(D-1)
2 dx
dx
帶回原式得
2
[ D(D-1) + D + 9λ ] y = [ D + 9λ ] y = 0
2 mt
令 λ = k , y = e
2 2 mt
[ D + 9k ] e = 0
2 2
m + (3k) = 0
m = ±3ki
y = c1 cos(3kt) + c2 sin(3kt)
= c1 cos(3klnx) + c2 sin(3klnx)
由 y'(1) = 0 知
-1 -1 ｜
- 3k x c1 sin(3klnx) + 3k x c2 cos(3klnx)｜ = 3k c2 = 0
｜x=1
因k≠0 , 得c2 = 0
由 y(e) = 0 知
c1 cos(3k) = 0
因c1≠0 (y≠0) , 得 3k = (n-1/2)π
2 2n-1 2
eigenvalues λn= kn = (------π)
6
eigenfunctions yn = bn cos[(n-1/2)πlnx]
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