Re: [理工] [工數]-二階ODE
※ 引述《osunriceo (語)》之銘言：
: ( x^2 - 1 )x^2y'' - (x^2 + 1)xy' + (x^2 + 1)y = 0
: 有勞高手幫忙
: 想用因式分解
: 但是不知道從何分起
: 想用因變式自變式的方法解
: 可是感覺這樣會做得落落長...
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(x^2 - 1)x^2y'' - (x^2 + 1)xy' + (x^2 + 1)y = 0
→ [ (x^2-1)xD - (x^2+1) ]( xD - 1 )y = 0
→ [ (x^2-1)xD - (x^2+1) ][(x^2)(y/x)'] = 0
x^3
→ (x^2-1)^2 [ ────(y/x)']' = 0 if x≠1
x^2 - 1
x^3
→ [ ────(y/x)']' = 0
x^2 - 1
→ (y/x)' = c1*(1/x - 1/x^3)
→ y/x = c1*[ln｜x｜ + 1/(2x^2)] + c2
or y = c1*[xln｜x｜ + 1/(2x)] + c2*x
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