# 11 let then and substitution into the ode results in

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online