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114    CHAPTER 3  The Laplace Transform

                                 Apply the Laplace transform to the differential equation to get
                                                                                         1
                                                    2


                                                   s Y(s) − sy(0) − y (0) + 2L[ty ](s) − 4Y(s) = .
                                                                                         s
                                 Now y(0) = y (0) = 0, and

                                                              d

                                                  L[ty ](s) =−   L[y ](s)
                                                              ds
                                                              d

                                                          =−    (sY(s) − y(0)) =−Y(s) − sY (s).
                                                              ds
                                 The transformed differential equation is therefore
                                                                                     1
                                                        2
                                                       s Y(s) − 2Y(s) − 2sY (s) − 4Y(s) =
                                                                                     s
                                 or

                                                                  3   s        1

                                                            Y +     −    Y =−    .
                                                                  s   2       2s  2
                                 This is a linear first order differential equation for Y. To find an integrating factor, first compute
                                                              3  s              1  2

                                                               −     ds = 3ln(s) − s .
                                                              s  2              4
                                 The exponential of this function is a integrating factor. This is
                                                                         2
                                                                   e 3ln(s)−s /4
                                    3 −s 4
                                 or s e  2  . Multiply the differential equation for Y by this to obtain
                                                                          1   2
                                                                 2
                                                              3 −s /4

                                                             (s e  Y) =− se  −s /4 .
                                                                          2
                                 Then
                                                                          2
                                                               3 −s /4
                                                              s e  2  Y = e −s /4  + c.
                                 Then
                                                                     1    c  s /4
                                                                            2
                                                               Y(s) =  +   e  .
                                                                     s  3  s  3
                                                                                       3
                                 In order to have lim s→0 Y(s) = 0, choose c = 0, obtaining Y(s) = 1/s . The solution is
                                                                       1
                                                                 y(t) = t .
                                                                         2
                                                                       2
                                 3.7.2 Bessel Functions
                                   If n is a nonnegative integer, the differential equation
                                                                            2
                                                             2
                                                                        2

                                                            t y + ty + (t − n )y = 0

                                   is called Bessel’s equation of order n.
                                 This is usually considered for t ≥ 0. Bessel’s equation is second order, and the phrase order
                                 n refers to the parameter n in the coefficient of y. Solutions of Bessel’s equation are called
                                 Bessel functions of order n, and they occur in many settings, including diffusion processes, flow
                                 of alternating current, and astronomy. Bessel functions and some applications are developed in
                                 detail in Section 15.3.






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                                   October 14, 2010  14:14  THM/NEIL    Page-114        27410_03_ch03_p77-120
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