Page 44 - LAPLACE TRANSFORM
P. 44

Example   4






               Find the Laplace Transforms using the Multiplication by       of
                                                                                              

               the followings:-

                                                   2as
                    Prove that  ( sin.L t  at )=
                                                  2
                                                (s + a 2 ) 2
                                                        a
                    We know that :  {sin atL     } =           n = 1
                                                    (s +  a 2 )
                                                      2

                                                               (−=     ) 1  d  {  2  a  2  }
                                                                ds  (s + a  )

                     *use Quotient Rule to solve this question

                          u  = a                        v =  s (  2  + a )
                                                                 2

                                du
                                                                        Formula Quotient Rule
                                                        dv
         Find the Laplace Transform using the Third Multiplication
                                0
                              =
                                                           =
                                                              s 2
                           ds
                                                        ds
                                               n
                                         of t  of the following :-              du  − u  dv
                                                                              v
                                                                             { et .L  2  at }  d  ds  ds
                                     2
                              d }  (s + a 2 )(  ) 0 − (a )( 2s )       ds  =       v 2
                              at =
                           2
                                          2
                               L { et .  ds    (s +  a 2 ) 2
                                 =   ) 0 (  − 2as                       e ,
                                                                         at
                                     2
                                   (s + a 2 ) 2                    f(t) =
                                     − 2as
                                 =
                                                               f(t)
                                                                     =
                                   (s + a 2 ) 2             Prove        e at
                                     2

                                       Therefore,
                                          − 2as              d  2   1  
                                 =
                                   (−
                                     =n 1 )[ (s +  a  2 ) 2  ] at }  = ( − )1  2    
                                        2,  { et .L 22
                                                             ds 2   − as  
                                ∴     2as
                                     2
                                   (s + a 2 ) 2                                2
                                                                         = )1(  d   1   
                                                                                
                                                                             ds 2   − as  



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