Chem Differential Eq HW Solutions Fall 2011 203

Chem Differential Eq HW Solutions Fall 2011 203 - Section...

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Section A.6 The Method of Frobenius A203 We now turn to the second solution: y = ky 1 ln x + ± m =0 b m x m - 2 ; y ± = ± 1 ln x + k y 1 x + ± m =0 ( m - 2) b m x m - 3 ; y ±± = ±± 1 ln x + 2 k x y ± 1 - k x 2 y 1 + ± m =0 ( m - 2)( m - 3) b m x m - 4 . Plug into xy ±± +(1 - x ) y ± + y =0 : 2 kxy ± 1 + ±± 1 x 2 ln x - 1 + ± m =0 ( m - 2)( m - 3) b m x m - 2 + 4 ± 1 x ln x +4 1 + ± m =0 4( m - 2) b m x m - 2 +(2 - x 2 ) ± 1 ln x +(2 - x 2 ) ± m =0 b m x m - 2 2 kxy ± 1 +3 1 + ± m =0 [( m +2) 2 - ( m b m +2 + b m ] x m ( m 2 - ( m b m +2 + b m Take k = 0 and for all m 0, b m +2 = - b m ( m + 2)( m +1) . b 0 = 1 and
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