Page 77 - LAPLACE TRANSFORM
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Example   1






              Solve the differential equation         dy  + y = 3  by  using  the  Inverse  Laplace
                                                      dt
              Transform. Given that y(0)= 0 when t=0.


               STEP 1: Take the Laplace Transform for both sides.

                          L [ ] )(ty '  +  y = L [ ] 3


               STEP 2: Substitude the theorem for Left Hand Side (LHS) .

                          Find the Laplace for Right Hand Side (RHS)
                                             3
                                    0
                          [sY ( s) −  y ) +  ] y =
                                    (
                                             s

               STEP 3 : Substitude the values of y(0)=1, t=0
                                         3
                          [sY( s) − 0 +  ] y =
                                         s
               STEP 4 : Solve for Y(s)
                                       3
                           [sY (s ) + y ] =
                                        s
                                       3
                           Y (s )[ +  ] 1 =
                                s
                                       s
                                          3
                                Y (s ) =
                                         s
                                       s ( +  ) 1
               STEP 5 : Solve for the inverse ,     (    ) =      −     [    (    )] using Partial Fraction.
                                        3   
                            y  ) (t  = L −1   
                                        s (s + )1  
                               3   =  A +   B
                            s (s +  ) 1  s  (s +  ) 1

                                      3 =  A (s +  ) 1 + B (s )  ………………equation 1
               If  s= 0

                                3 = A  0 ( +  ) 1 +  ) 0 ( B
                                3 =    ) 1 ( A

                                              A =  3
               If s= -1
                               3 = A (− 1+  ) 1 + B (−  ) 1

                               3 = B (−  ) 1
                                 B  =  − 3





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