Series Solutions of Differential Equations

Series Solutions of Differential Equations - 930 CHAPTER 17...

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930 CHAPTER 17 Ordinary Differential Equations 17.7 Series Solutions of Differential Equations In Section 17.5 we developed a recipe for solving second-order, linear, homogeneous differential equations with constant coefficients: ay ±± + by ± + cy = 0 and Euler equations of the form ax 2 y ±± + bxy ± + = 0 . Many of the second-order, linear, homogeneous differential equations that arise in applications do not have constant coefficients and are not of Euler type. If the coefficient functions of such an equation are sufficiently well-behaved, we can often find solutions in the form of power series (Taylor series). Such series solutions are frequently used to define new functions, whose properties are deduced partly from the fact that they solve particular differential equations. For example, Bessel functions of order ν are defined to be certain series solutions of Bessel’s differential equation x 2 y ±± + xy ± + ( x 2 ν 2 ) y = 0 . Series solutions for second-order homogeneous linear differential equations are most easily found near an ordinary point of the equation. This is a point x = a such that the equation can be expressed in the form y ±± + p ( x ) y ± + q ( x ) y = 0 , where the functions p ( x ) and q ( x ) are analytic at x = a . (Recall that a function f is analytic at x = a if f ( x ) can be expressed as the sum of its Taylor series in powers of x a in an interval of positive radius centred at x = a .) Thus we assume p ( x ) = ± n = 0 p n ( x a ) n , q ( x ) = ± n = 0 q n ( x a ) n , with both series convergingin some interval of the form a R < x < a + R . Frequently p ( x ) and q ( x ) are polynomials so are analytic everywhere. A change of independent variable ξ = x a will put the point x = a at the origin ξ = 0, so we can assume that a = 0. The following example illustrates the technique of series solution around an ordi- nary point. Example 1 Find two independent series solutions in powers of x for the Hermite equation y ±± 2 ± + ν y = 0 . For what values of ν does the equation have a polynomial solution? Solution We try for a power series solution of the form y = ± n = 0 a n x n = a 0 + a 1 x + a 2 x 2 + a 3 x 3 +··· , so that y ± = ± n = 1 na n x n 1 y ±± = ± n = 2 n ( n 1 ) a n x n 2 = ± n = 0 ( n + 2 )( n + 1 ) a n + 2 x n .
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SECTION 17.7: Series Solutions of Differential Equations 931 (We have replaced n by n + 2inordertoget x n in the sum for y 00 .) We substitute these expressions into the differential equation to get X n = 0 ( n + 2 )( n + 1 ) a n + 2 x n 2 X n = 1 na n x n + ν X n = 0 a n x n = 0 or 2 a 2 + ν a 0 + X n = 1 h ( n + 2 )( n + 1 ) a n + 2 ( 2 n ν) a n i x n = 0 . This identity holds for all x provided that the coefFcient of every power of x vanishes; that is, a 2 =− ν a 0 2 , a n + 2 = ( 2 n a n ( n + 2 )( n + 1 ) ,( n = 1 , 2 , ··· ).
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This note was uploaded on 03/21/2012 for the course STAT 101 taught by Professor Graham during the Spring '08 term at Iowa State.

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Series Solutions of Differential Equations - 930 CHAPTER 17...

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