B&D Ch 5_2 Power series example 9_15_11.docx

B&D Ch 5_2 Power series example 9_15_11.docx - Prof....

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1 Prof. S.L. Phoenix, 9/15/11 B&D Section 5.2: An Example We use the power series solution method, expanding about 0 xa , to solve the differential equation 0 yx y y   ( 1 ) subject to the initial conditions   02 , 03 yy  ( 2 ) Here power series solution method means we are going to seek a solution of the general form    0 n n n a x a  (3a) and since 0 a this reduces to  0 n n n ax ( 3 b ) Note: OK, why a power series? Well equation (1) is a 2 nd order linear ordinary differential equation with at least one non-constant coefficient (the coefficient on y is x ) so we can rest assured that our previous methods of Chapter 3 were not going to work – so we turn to the power series approach. We do know, however, that there must be two linearly independent solutions, which can call 1 y and 2 y , and this could raise anxiety since so far we are working with one power series, but let’s see what happens. To substitute the series (3b) into (1) we need 11 10 nn y x na x x na x   ( 4 ) and 2 2 1 n n n n n a x ( 5 ) and substituting into (1) gives 2 20 0 n n n n  ( 6 ) By so-called “shifting indices” in the first summation (i.e., add 2 to each n inside the sum and subtract 2 from n in the lower limit) we get 2 00 0 21 0 n n n a x n a x a x  ( 7 )
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2 and through factoring out n x in each term we get   2 0 21 1 0 n nn n a n a x  ( 8 ) Since x
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B&D Ch 5_2 Power series example 9_15_11.docx - Prof....

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