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4.2 Frobenius Solutions  129


                                           Suppose that the indicial equation has real roots r 1 and r 2 with r 1 ≥ r 2 . Then the following
                                        conclusions hold.
                                        (2) If r 1 − r 2 is not a positive integer, then there are two linearly independent Frobenius
                                        solutions
                                                                   ∞                     ∞

                                                                                            ∗
                                                            y 1 (x) =  c n x n+r 1  and  y 2 (x) =  c x  n+r 2
                                                                                            n
                                                                  n=0                   n=0
                                        with c 0  = 0 and c  = 0. These solutions are valid at least in an interval (0,h) or (−h,0).
                                                      ∗
                                                      0
                                        (3) If r 1 −r 2 = 0, then there is a Frobenius solution
                                                                              ∞

                                                                       y 1 (x) =  c n x n+r 1
                                                                             n=0
                                        with c 0  = 0, and there is a second solution
                                                                                  ∞

                                                                                     ∗
                                                                  y 2 (x) = y 1 ln(x) +  c x  n+r 1 .
                                                                                     n
                                                                                 n=1
                                        These solutions are linearly independent on some interval (0,h).
                                        (4) If r 1 −r 2 is a positive integer, then there is a Frobenius solution
                                                                              ∞

                                                                      y 1 (x) =  c n x  n+r 1 .
                                                                             n=0
                                        with c 0  = 0, and there is a second solution
                                                                                    ∞

                                                                                       ∗
                                                                 y 2 (x) = ky 1 (x)ln(x) +  c x  n+r 2
                                                                                       n
                                                                                   n=0
                                        with c  = 0. y 1 and y 2 are linearly independent solutions on some interval (0,h).
                                             ∗
                                             0
                                           The method of Frobenius consists of using Frobenius series and Theorem 4.2 to solve equa-
                                        tion (4.7) in some interval (−h,h), (0,h),or (−h,0), assuming that 0 is a regular singular point.
                                        Proceed as follows:


                                                             ∞    n+r
                                        Step 1. Substitute y =  c n x  into the differential equation, and solve for the roots r 1 and
                                                             n=0
                                              r 2 of the indicial equation for r. This yields a Frobenius solution (which may or may not
                                              be a power series).
                                        Step 2. Depending on which of Cases (2), (3), or (4) of Theorem 4.2 applies, the theorem pro-
                                              vides a template for a second solution which is linearly independent from the first.
                                              Once we know what this second solution looks like, we can substitute its general
                                              form into the differential equation and solve for the coefficients and, in Case (4), the
                                              constant k.

                                           We will illustrate the Cases (2), (3), and (4) of the Frobenius theorem. For case (2), Exam-
                                        ple 4.5, we will provide all of the details. In Cases (3) and (4) (Examples 4.6, 4.7, and 4.8), we
                                        will omit some of the calculations and include just those that relate to the main point of that
                                        case.






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                                   October 14, 2010  14:17  THM/NEIL   Page-129        27410_04_ch04_p121-136
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