Re: [工數]一題卡很久的高階ODE...
忍不住就想提供另一做法(因為想很久也解很久)XD
a a
y'' - ------ y = 0 <=> xy'' - ----- y = 0
2 x
x
Let u = y/x => y' = xu' + u , y'' = xu''+ 2u'
=> x(xu'' + 2u') - au = 0
2 ''
=> x u + 2xu' - au = 0 is a Cauchy-Euler Equation
Let v = ln x. Then,
n n
2 2 2 n d n d
xD u = D u and x D u = D u - D u, where D = ---- and D = ----
x v x v v x n v n
dx dv
2
=> D u + D u - au = 0
v v
-1 ±√(1+4a)
The characteristic roots are ------------------
2
-1 + √(1+4a) -1 -√(1+4a)
Let r1 = --------------, r2 = -------------
2 2
r1 r2
=> u = c1 x + c2 x
r1+1 r2+1
=> y = xu = c1 x + c2 x
By x = kR, y = 0 and x = R, y = w,
r1+1 r2+1
[ (kR) (kR) ] [ c1 ] [ 0 ]
[ ] [ ] = [ ]
[ r1+1 r2+1 ] [ ] [ ]
[ R R ] [ c2 ] [ w ]
r1+1 r2+1 r1+r2+2
Det = (k - k ) R
r1 + r2 + 2 = -1 + 2 = 1
r2+1 r2+1
[ c1 ] 1 [ R -(kR) ] [ 0 ]
[ ] = ----------------- [ ] [ ]
[ ] r1+1 r2+1 [ r1+1 r1+1 ] [ ]
[ c2 ] R(k - k ) [ - R (kR) ] [ w ]
r2+1 r2
- (kR) R
=> c1 = ----------------- w = --------------- w
r1+1 r2+1 r1-r2
R(k - k ) (1 - k )
r1+1 r1
(kR) R
c2 = ------------------ w = --------------- w
r1+1 r2+1 r2-r1
R(k - k ) (1 - k )
r1-r2 = √(1+4a)
1 + √(1+4a) 1 - √(1+4a)
r1+1 = ------------------, r2+1 = ---------------
2 2
So, -[1+√(1+4a)]/2 [-1+√(1+4a)]/2
R [1+√(1+4a)]/2 R [1-√(1+4a)]
y = w{------------------- x + -------------------- x }
√(1+4a) -√(1+4a)
1 - k 1 - k
--
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