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3.4 Convolution    101





                                                         25


                                                         20


                                                         15


                                                         10



                                                         5


                                                         0
                                                           0       2       4       6       8      10
                                                                               t
                                                         FIGURE 3.21 Replacement function.


                                           Figure 3.21 is a graph of this replacement function for A = 2, B = 0.001, and k = 1. As
                                        expected, r(t) is a strictly increasing function, because it measures total replacements up to a
                                        given time. The graph gives an indication of how the drug needs to be replenished to maintain
                                        f (t) doses at time t.



                               SECTION 3.4        PROBLEMS


                            In each of Problems 1 through 8, use the convolution the-  In each of Problems 9 through 16, use the convolution
                            orem to help compute the inverse Laplace transform of  theorem to write a formula for the solution in terms of f .
                            the function. Wherever they occur, a and b are positive



                            constants.                                     9. y − 5y + 6y = f (t); y(0) = y (0) = 0
                                    1                                      10. y + 10y + 24y = f (t); y(0) = 1, y (0) = 0



                            1.
                               (s + 4)(s − 4)                              11. y − 8y + 12y = f (t); y(0) =−3, y (0) = 2
                                      2
                                2



                                 1


                            2.      e −2s                                  12. y − 4y − 5y = f (t); y(0) = 2, y (0) = 1

                               2
                               s + 16


                                     s                                     13. y + 9y = f (t); y(0) =−1, y (0) = 1
                            3.
                                          2
                               (s + a )(s + b )                            14. y − k y = f (t); y(0) = 2, y (0) =−4
                                    2
                                      2
                                                                                  2
                                2


                                   s  2                                    15. y (3)  − y − 4y + 4y = f (t); y(0) = y (0) =



                            4.
                                     2
                               (s − 3)(s + 5)                                 1, y (0) = 0

                                  1                                            (4)
                            5.                                             16. y  − 11y + 18y = f (t); y(0) = y (0) = y (0) =
                                 2
                                    2 2
                                                                               (3)
                               s(s + a )                                      y (0) = 0
                                 1
                            6.
                               s (s − 5)                                   In each of Problems 17 through 23, solve the integral
                               4
                                                                           equation.
                                 1
                            7.       e −4s
                               s(s + 2)                                                
  t
                                                                           17. f (t) =−1 +  f (t − τ)e −3τ  dτ
                                  2                                                     0
                            8.                                                         
  t  f (t − τ)sin(τ)dτ
                                 2
                               2
                               s (s + 5)                                   18. f (t) =−t +  0
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