lecture12 - Differential Equations& Linear Algebra...

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Unformatted text preview: Differential Equations & Linear Algebra Lecture 12 Jianli XIE Email: [email protected] Department of Mathematics, SJTU Fall 2010 Contents Series solutions near an ordinary point Contents Series solutions near an ordinary point Brief review for midterm Example Example Find a series solution in powers of x of the equation y 00- xy = 0 ,-∞ < x < ∞ . Solution Since every point is an ordinary point, we assume that y = ∞ X n =0 a n x n . Then y 00 = ∞ X n =2 n ( n- 1) a n x n- 2 = ∞ X n =0 ( n + 2)( n + 1) a n +2 x n . Substituting these two series into the equation, we obtain Example ∞ X n =0 ( n + 2)( n + 1) a n +2 x n = x ∞ X n =0 a n x n = ∞ X n =0 a n x n +1 . To compare the coefficients of same powers of x , we shift index of summation: 2 a 2 + ∞ X n =1 ( n + 2)( n + 1) a n +2 x n = ∞ X n =1 a n- 1 x n . This implies a 2 = 0 and the recurrence relation ( n + 2)( n + 1) a n +2 = a n- 1 , n = 1 , 2 , 3 , ··· . Since a n +2 is given in terms of a n- 1 , the coefficients are determined in steps of three. Example That is, a determines a 3 , a 6 , ··· ; a 1 determines a 4 , a 7 , ··· ; and a 2 determines a 5 , a 8 , ··· . Since a 2 = 0 , we immediately conclude that a 5 = a 8 = a 11 = ··· = 0 . For a 3 , a 6 , a 9 , ··· we put n = 1 , 4 , 7 , ··· in the recurrence relation to get a 3 = a 2 · 3 , a 6 = a 2 · 3 · 5 · 6 , a 9 = a 2 · 3 · 5 · 6 · 8 · 9 , ··· . This suggests the general formula a 3 n = a 2 · 3 · 5 · 6 ··· (3 n- 1)(3 n ) , n = 1 , 2 , ··· . Example For a 4 , a 7 , a 1 , ··· we put n = 2 , 5 , 8 , ··· in the recurrence relation to get a 4 = a 1 3 · 4 , a 7 = a 1 3 · 4 · 6 · 7 , a 10 = a 1 3 · 4 · 6 · 7 · 9 · 10 , ··· In general, a 3 n +1 = a 1 3 · 4 · 6 · 7 ··· (3 n )(3 n + 1) , n = 1 , 2 , ··· . Thus the general solution is y = a 1 + x 3 2 · 3 + x 6 2 · 3 · 5 · 6 + ··· + x 3 n 2 · 3 ··· (3 n- 1)(3 n ) + ··· + a 1 x + x 4 3 · 4 + x 7 3 · 4 · 6 · 7 + ··· + x 3 n +1 3 · 4 ··· (3 n )(3 n +1) + ··· . Example It is easy to show by the ratio test that both series have radius of convergence ρ = ∞ . Let y 1 and y 2 denote the functions to which the two series converge, respectively. Notice that y 1 satisfies the initial conditions y 1 (0) = 1 , y 1 (0) = 0 and that y 2 satisfies the initial conditions y 2 (0) = 0 , y 2 (0) = 1 ....
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This note was uploaded on 05/17/2011 for the course ME 255 taught by Professor Jianlixie during the Fall '10 term at Shanghai Jiao Tong University.

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lecture12 - Differential Equations& Linear Algebra...

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