The zeros of the roots of and the roots of are at are

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Unformatted text preview: — CHAPTER 5. —— 15. If and are solutions, then substituting into the ODE results in . Setting , we find that . Similarly, substituting into the ODE results in . Therefore and may not be analytic. If they were, Theorem would guarantee that and were the only two solutions. But note that an arbitrary value of cannot be a linear combination of and Hence must be a singular point. 16. Let . Substituting into the ODE, That is, Setting the coefficients equal to zero, we obtain for is The coefficient . It is easy to see that . Therefore the general solution , which can be arbitrary. 17. Let . Substituting into the ODE, That is, Combining the series, we have ________________________________________________________________________ page 198 ————————————————————————— CHAPTER 5. —— Setting the coefficient equal to zero, Note that the indices differ by two, so for and for . and Hence the general solution is . The coefficient , which can be arbitrary. 19. 21 Let . Substituting into the ODE, That is, Combining the series, we have Setting the coefficients equal to zero, Hence the general solution is The coefficient 2 191. Let and for . , which can be arbitrary. . Substituting into the ODE, ________________________________________________________________________ page 199 —————————————————————————— CHAPTER 5. —— That is, Combining the series, and the nonhomogeneous terms, we have Setting the coefficients equal to zero, we obtain , , , and . The indices differ by two, so for , and for Hence the general solution is Collecting the terms containing , Upon inspection, we find that Note that the given ODE is first order linear, with integrating factor general solution is given by . The ________________________________________________________________________ page 200 —————————————————————————— CHAPTER 5. —— 23. If If 24 , then . If , then . Based on Prob. for , then for . As a result, . As a result, , Normalizing the polynomials, we obtain Similarly, ________________________________________________________________________ page 201 —————————————————————————— CHAPTER 5. —— . . has no roots. The zeros of by has one root at are , . The zeros of . The roots of , and . The roots of , are at are given are given by . 25. Observe that . But for all nonnegative integers . 27. We have , which is a polynomial of degree . Differentiating n times, , in which the lower index is . Note that if , then . ________________________________________________________________________ page 202 —————————————————————————— CHAPTER 5. —— Now shift the index, by setting . Hence Based on Prob. , 29. Since the polynomials , , of is , any polynomial, , of degree , are linearly independent, and the degree can be expressed as a linear combin...
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This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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