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Unformatted text preview: Chap . 5 Series Solutions of ODEs Special Functions This chapter introduces you to higher functions that are needed in physics and its engineering applications. Just as you could not do problems in calculus without knowing trigonometric and exponential functions, you could not solve important models involving ordinary or partial differential equations without knowing Bessel functions and Legendre polynomials, to mention just the two most prominent classes of these functions. Although these and the many other special functions of practical interest have quite different properties and serve very distinct purposes, it is most remarkable that these functions are accessible by the same instrument, namely, power series , (whose simplest examples you know from calculus), perhaps multiplied by a fractional power or a logarithm. You should see and learn which of the many functions and series are of importance in applications and what methods are available for investigating properties through series and for discovering relations between functions, in addition to the calculation of values by series and representing the results graphically by curves. Such an overview will help you to find your way through the vast number of books and articles in journals related to special functions when you need help in solving specific engineering or other problems. Your CAS knows all the functions that you will ever need, but it is your task to establish a general orientation in this wide field of relationships and formulas, in order to find and select what functions and relations you need for specific tasks and situations. Sec . 5 . 1 Power Series Method Problem Set 5 . 1 . Page 170 5 . Terminating power series . The ODE u 2 u x U y U ¡ y can be solved by separating variables, dy y ¡ dx x u 2 , ln  y  ¡ ln  x u 2  u c , y ¡ c ¢ u x u 2 U . The solution is a polynomial, a terminating power series, and you will see why, and how termination occurs. Substitute the power series for y and y U into the given ODE to get u 2 u x Uu a 1 u 2 a 2 x u 3 a 3 x 2 u ¡U ¡ a u a 1 x u a 2 x 2 u ¡ . Write the left side as a power series by multiplying out and ordering, 2 a 1 u 4 a 2 x u 6 a 3 x 2 u ¡ u a 1 x u 2 a 2 x 2 u 3 a 3 x 3 u ¡ ¡ 2 a 1 u u 4 a 2 u a 1 U x u u 6 a 3 u 2 a 2 U x 2 u ¡ . Now compare powers on both sides: a ¡ 2 a 1 hence a 1 ¡ a 2 a 1 ¡ 4 a 2 u a 1 a 2 ¡ a 2 ¡ 6 a 3 u 2 a 2 a 3 ¡ 0, etc. and you see how it happens that the power series terminates after two terms. a remains arbitrary, so that you have obtained a general solution of your firstorder ODE, namely, y ¡ a u 1 u x 2 U . 13 . Verhulst ODE , initial value . y U ¡ y u y 2 ¡ y u 1 u y U has the general solution (see (9) on p. 31 in Sec....
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.
 Spring '12
 M.lee

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