11 let then and substitution into the ode results in

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Unformatted text preview: ________________________________________________________________________ page 179 —————————————————————————— CHAPTER 5. —— . Setting the coefficients equal to zero, we obtain , , and , Observe that for two, we also have and for . , we obtain . Since the indices differ by . Therefore the general solution is a polynomial . Hence the linearly independent solutions are and 10. Let . . Then Substitution into the ODE results in First write . It follows that . We obtain , and , . Note that for , . Since the indices differ by two, we also have . On the other hand, for , for Therefore the general solution is ________________________________________________________________________ page 180 —————————————————————————— CHAPTER 5. —— . Hence the linearly independent solutions are and . 11. Let . Then and Substitution into the ODE results in . Before proceeding, write and It follows that . We obtain , , and , . The indices differ by two, so for and ________________________________________________________________________ page 181 —————————————————————————— CHAPTER 5. —— Hence the linearly independent solutions are 12. Let . Then and Substitution into the ODE results in . Before proceeding, write and It follows that . We obtain for and . Writing out the individual equations, ________________________________________________________________________ page 182 —————————————————————————— CHAPTER 5. —— The coefficients can be calculated successively as , , , . We can now see that for , is proportional to . In fact, for , . Therefore the general solution is . Hence the linearly independent solutions are 13. Let and . Then and Substitution into the ODE results in . First write We then obtain It follows that and ________________________________________________________________________ page 183 —————————————————————————— CHAPTER 5. —— for . The indices differ by two, so for and Hence the linearly independent solutions are 15 . From Prob. , we have and Since and , we have . . That is, . The four- and five-term polynomial approximations are ________________________________________________________________________ page 184 —————————————————————————— CHAPTER 5. —— . . The four-term approximation on the interval . 17 Since appears to be reasonably accurate within % . From Prob. , the linearly independent solutions are and , we have . That is, . The four- and five-term polynomial approximations are ________________________________________________________________________ page 185 —————————————————————————— CHAPTER 5. —— . . The four-term approximation on the interval . 18 . From Prob. appears to be reasonably accurate within , we have and Since an...
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