L12 - Series SolutionsMotivatio So far the situation is...

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Unformatted text preview: Series SolutionsMotivatio So far the situation is this: if we have one solution y 1 of y 00 + p ( x ) y + q ( x ) y = 0 ( H ) we can, with luck in integration, parlay it to a FS { y 1 ,y 2 } ; find, with luck in in- tegration, a particular solution Y of the inhomogeneous equation y 00 + p ( x ) y + q ( x ) y = g ( x ) , ( NH ) which then gives us the general solution of ( NH ) as y = Y + c 1 y 1 + c 2 y 2 . OK, so where does the first solution y 1 come from? Here is an idea: Often the coefficients p ( x ) and q ( x ) of ( H ) are polynomials, so we might try for a solution that is also a poly- nomial. Example: Consider the SOLODE y 00- xy = 0 . ( A ) Here p ( x ) = 0 and q ( x ) =- x , both polyno- mials. We try y ( x ) = a + a 1 x + a 2 x 2 + + a N x N = N X n =0 a n x n with a N 6 = 0 ( N is the index of the last nonzero coefficient.) 2 y = a + a 1 x + a 2 x 2 + + a N x N y 00 =2 a 2 + 6 a 3 x N ( N- 2) a N x N- 2 xy = a x + + a N- 1 x N + a N x N +1 L [ y ] = 2 a 2 +(6 a 3- a ) x + - a N x N +1 L [ y ] = 0 implies a N = 0 : a contradiction! Trying for a polynomial solution leads nowher What to do? 3 y = a + a 1 x + a 2 x 2 + + a N x N y 00 =2 a 2 + 6 a 3 x N ( N- 2) a N x N- 2 xy = a x + + a N- 1 x N + a N x N +1 L [ y ] = 2 a 2 +(6 a 3- a ) x + - a N x N +1 L [ y ] = 0 implies a N = 0 : a contradiction! Trying for a polynomial solution leads nowher What to do? The trouble is with the last term a N x N of the polynomial; if it werent there the problem would not arise. The mathemati- cian (recall: thats a lazy person with in- testinal fortitude) would try for a polyno- mial without a last term: a power series. 3 Review of Infinite Series Everyone knows how to sum finitely many numbers , 1 , 2 ,..., N : + 1 + + N = N X n- 1 n is well, you just sum....
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L12 - Series SolutionsMotivatio So far the situation is...

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