lecture12 - Dierential Equations Linear Algebra Lecture 12...

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Differential Equations & Linear Algebra Lecture 12 Jianli XIE Email: [email protected] Department of Mathematics, SJTU Fall 2010
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Contents Series solutions near an ordinary point
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Contents Series solutions near an ordinary point Brief review for midterm
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Example Example Find a series solution in powers of x of the equation y - xy = 0 , -∞ < x < . Solution Since every point is an ordinary point, we assume that y = n =0 a n x n . Then y = n =2 n ( n - 1) a n x n - 2 = n =0 ( n + 2)( n + 1) a n +2 x n . Substituting these two series into the equation, we obtain
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Example n =0 ( n + 2)( n + 1) a n +2 x n = x n =0 a n x n = n =0 a n x n +1 . To compare the coefficients of same powers of x , we shift index of summation: 2 a 2 + n =1 ( n + 2)( n + 1) a n +2 x n = 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.
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Example That is, a 0 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 0 2 · 3 , a 6 = a 0 2 · 3 · 5 · 6 , a 9 = a 0 2 · 3 · 5 · 6 · 8 · 9 , · · · . This suggests the general formula a 3 n = a 0 2 · 3 · 5 · 6 · · · (3 n - 1)(3 n ) , n = 1 , 2 , · · · .
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Example For a 4 , a 7 , a 1 0 , · · · 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 0 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) + · · · .
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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 . Thus W ( y 1 , y 2 )(0) = 1 = 0 , and consequently y 1 and y 2 are linearly independent. Hence the general solution is y = a 0 y 1 ( x ) + a 1 y 2 ( x ) , < x < .
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Example Example Find a series solution in powers of x - 1 of the equation y - xy = 0 , -∞ < x < .
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