Page 204 - Numerical Analysis Using MATLAB and Excel
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Summary


                                            yt() =  y  Natural  +  y  Forced  =  y +  y F
                                                                          N
                                                    Response   Response
                  The natural response  y N   contains arbitrary constants and these can be evaluated from the

                  given initial conditions. The forced response y F  , however, contains no arbitrary constants. It is
                  imperative to remember that the arbitrary constants of the natural response must be evaluated
                  from the total response.


               • For an nth   order homogeneous differential equation the solutions are
                                            s t          s t         s t             s t
                                    y =  k e ,  1  1  y =  k e ,  2  2  y =  k e ,  3  3  … ,  y =  k e  n
                                      1
                                                                              n
                                                              3
                                                  2
                                                                                   n
                           ,
                                 ,
                              ,
                  where s s … s    n  are the solutions of the characteristic equation
                          1
                             2
                                                n
                                             a s + a n –  1 s n –  1  +  … +  a s +  a 0  =  0
                                              n
                                                                    1
                                 ,
                  and a a,  n  n –  1 ,  … a a 0  are the constant coefficients of the ODE
                                    ,
                                   1
               • If the roots of the characteristic equation are distinct, the   solutions of the natural response
                                                                           n
                  are independent and the most general solution is:
                                                        s t    s t         s t
                                               y =  k e  1  +  k e  2  +  … +  k e  n
                                                      1
                                                             2
                                                 N
                                                                         n
               • If the solution of the characteristic equation contains   equal roots, the most general solution
                                                                      m
                  has the form:
                                                                 s t      s t         s t
                                                                  1
                                    y =  (  k +  k t + … +  k t m –  1 )  e  +  k n –  i e  2  +  … +  k e  n
                                               2
                                                         m
                                     N
                                           1
                                                                                    n
               • If the characteristic equation contains complex roots, these occur as complex conjugate pairs.
                                                                   β
                                                            α
                  Thus, if one root is s =  –  α +  jβ  where   and   are real numbers, then another root is
                                        1
                   s =  – α jβ . Then, for two complex conjugate roots we evaluate the constants from the
                          –
                    2
                  expressions
                                     s t    s t    – αt                    – αt
                                      1
                                                                       )
                                             2
                                  k e  +  k e  =  e  (  k cos βt + k sin βt =  e  k cos (  βt +  ϕ )
                                                                               5
                                   1
                                          2
                                                       3
                                                                 4
               • The forced response of a non−homogeneous ODE can be found by the method of undeter-
                  mined coefficients or the method of variation of parameters.
               • With the method of undetermined coefficients, the forced response is a function similar to the
                  right side of the non−homogeneous ODE. The form of the forced response for second order
                  non−homogeneous ODEs is given in Table 5.1.
               • In certain non−homogeneous ODEs, the right side ft()    cannot be determined by the method
                   of undetermined coefficients. For these ODEs we must use the method of variation of parame-
               Numerical Analysis Using MATLAB® and Excel®, Third Edition                              5−43
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