ch05 - CHAPTER 5 Series Solutions of ODEs. Special...

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Unformatted text preview: CHAPTER 5 Series Solutions of ODEs. Special Functions Changes of Text Minor changes, streamlining the text to make it more teachable, retaining the general structure of the chapter and the opportunity to familiarize the student with an overview of some of the techniques used in connection with higher special functions, also in connection with a CAS. SECTION 5.1. Power Series Method, page 167 Purpose. A simple introduction to the technique of the power series method in terms of simple examples whose solution the student knows very well. SOLUTIONS TO PROBLEM SET 5.1, page 170 2. y 5 a (1 2 1 _ 2 x 2 1 1 _ 8 x 4 2 1 • • • ) 5 a e 2 x 2 /2 4. A general solution is y 5 c 1 e x 1 c 2 e 2 x . Both functions of the basis contain every power of x . The power series method automatically gives a general solution in which one function of the basis is even and the other is odd, y 5 a (1 1 1 _ 2 x 2 1 _ 1 24 x 4 1 • • • ) 1 a 1 ( x 1 1 _ 6 x 3 1 _ 1 120 x 5 1 • • • ) 5 a cosh x 1 a 1 sinh x . 6. y 5 a (1 2 3 x 1 9 _ 2 x 2 2 _ 11 2 x 3 1 _ 51 8 x 4 2 1 • • • ). Even if the solution turns out to be a known function, as in the present case, y 5 a exp ( 2 3 x 2 x 3 ) as obtained by separating variables, it is generally not easy to recognize this from a power series obtained. Of course, this is generally not essential, because the method is primarily designed for ODEs whose solutions define new functions. The task of recognizing a power series as a known function does however occur in practice, for instance, in proving that certain Bessel functions or hypergeometric functions reduce to familiar known functions (see Secs. 5.4, 5.5). 8. Substitution of the power series for y and y 9 gives ( x 5 1 4 x 3 )( a 1 1 2 a 2 x 1 3 a 3 x 2 1 • • • ) 5 4 a 1 x 3 1 8 a 2 x 4 1 (12 a 3 1 a 1 ) x 5 1 • • • ) 5 (5 x 4 1 12 x 2 )( a 1 a 1 x 1 a 2 x 2 1 • • • ) 5 12 a x 2 1 12 a 1 x 3 1 (12 a 2 1 5 a ) x 4 1 (12 a 3 1 5 a 1 ) x 5 1 • • • ) Comparing coefficients of each power of x , we obtain x 2 : a 5 x 3 : 4 a 1 5 12 a 1 , a 1 5 x 4 : 8 a 2 5 12 a 2 1 5 a , a 2 5 x 5 : 1 2 a 3 1 a 1 5 12 a 3 1 5 a 1 , a 3 arbitrary x 6 : 1 6 a 4 1 2 a 2 5 12 a 4 1 5 a 2 , a 4 5 x 7 : 3 a 3 1 20 a 5 5 5 a 3 1 12 a 5 , a 5 5 1 _ 4 a 3 x 8 : 4 a 4 1 24 a 6 5 5 a 4 1 12 a 6 , 2 4 a 6 5 12 a 6 , a 6 5 0, etc. 97 im05.qxd 9/21/05 11:10 AM Page 97 Here we see why a power series may terminate in some cases. The answer is y 5 a 3 ( x 3 1 1 _ 4 x 5 ). 10. y 5 a ( 1 2 x 2 2 x 4 2 x 6 2 x 8 2 • • • ) 1 a 1 x 12. s 5 x 1 1 _ 3 x 3 1 _ 2 15 x 5 , s ( 1 _ 4 p ) 5 0.98674, a good approximation of the exact value 1 5 tan 1 _ 4 p . Through these calculations of values the student should realize that a series can be used for numeric work just like a solution formula, with proper caution regarding convergence and accuracy....
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ch05 - CHAPTER 5 Series Solutions of ODEs. Special...

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