Module

for

Laplace Transforms of Derivatives and Integrals

 

Chapter 12  Fourier Series and the Laplace Transform

12.6  Laplace Transforms of Derivatives and Integrals

    This section is a continuation of our development of the Laplace Transform in Section 12.5.  

 

Theorem 12.13  (Differentiation of f(t) ). Let [Graphics:Images/LaplaceDiffIntegrateMod_gr_1.gif] be continuous for [Graphics:Images/LaplaceDiffIntegrateMod_gr_2.gif], and of exponential order. Then

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_3.gif],  

where [Graphics:Images/LaplaceDiffIntegrateMod_gr_4.gif].   

Proof.

 

Corollary 12.1  (Differentiation of f(t) ).  If  [Graphics:Images/LaplaceDiffIntegrateMod_gr_5.gif]  are of exponential order, then  

        [Graphics:Images/LaplaceDiffIntegrateMod_gr_6.gif].  

Proof.

 

Example 12.13.  Show that  [Graphics:Images/LaplaceDiffIntegrateMod_gr_7.gif].

Solution.

If we let  [Graphics:Images/LaplaceDiffIntegrateMod_gr_8.gif], then [Graphics:Images/LaplaceDiffIntegrateMod_gr_9.gif]  and  [Graphics:Images/LaplaceDiffIntegrateMod_gr_10.gif],  and [Graphics:Images/LaplaceDiffIntegrateMod_gr_11.gif] is to be determined.  Because   [Graphics:Images/LaplaceDiffIntegrateMod_gr_12.gif],  we have  [Graphics:Images/LaplaceDiffIntegrateMod_gr_13.gif],  and Theorem 12.13 implies that

        [Graphics:Images/LaplaceDiffIntegrateMod_gr_14.gif]  

Thus, we have  [Graphics:Images/LaplaceDiffIntegrateMod_gr_15.gif],  and  [Graphics:Images/LaplaceDiffIntegrateMod_gr_16.gif]  from which it follows that

        [Graphics:Images/LaplaceDiffIntegrateMod_gr_17.gif].

Explore Solution 12.13.

 

Theorem 12.14  (Integration of f(t) ).  Let [Graphics:Images/LaplaceDiffIntegrateMod_gr_27.gif] be continuous for [Graphics:Images/LaplaceDiffIntegrateMod_gr_28.gif], and of exponential order and [Graphics:Images/LaplaceDiffIntegrateMod_gr_29.gif], then  

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_30.gif].

Proof.

 

Example 12.14.    (a)  Show that  [Graphics:Images/LaplaceDiffIntegrateMod_gr_31.gif],  (b)  Show that  [Graphics:Images/LaplaceDiffIntegrateMod_gr_32.gif].  

Solution.

Part (a). Use the fact that  [Graphics:Images/LaplaceDiffIntegrateMod_gr_33.gif],  and apply Theorem 12.14, with  [Graphics:Images/LaplaceDiffIntegrateMod_gr_34.gif]   and  [Graphics:Images/LaplaceDiffIntegrateMod_gr_35.gif],
then obtain

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_36.gif].


Part (b).  Now we can use this first result as a fact,  [Graphics:Images/LaplaceDiffIntegrateMod_gr_37.gif].  This time we apply Theorem 12.14, with  [Graphics:Images/LaplaceDiffIntegrateMod_gr_38.gif]   and  [Graphics:Images/LaplaceDiffIntegrateMod_gr_39.gif],
and obtain

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_40.gif].

Explore Solution 12.14 (a).

Explore Solution 12.14 (b).

 

    One of the main uses of the Laplace transform is its role in the solution of differential equations.  The utility of the Laplace transform lies in the fact that the transform of the derivative [Graphics:Images/LaplaceDiffIntegrateMod_gr_63.gif] corresponds to multiplication of the transform [Graphics:Images/LaplaceDiffIntegrateMod_gr_64.gif] by s and then the subtraction of [Graphics:Images/LaplaceDiffIntegrateMod_gr_65.gif].  This permits us to replace the calculus operation of differentiation with simple algebraic operations on transforms.

    This idea is used to develop a method for solving linear differential equations with constant coefficients.  Let's consider the initial value problem  

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_66.gif],

with initial conditions  [Graphics:Images/LaplaceDiffIntegrateMod_gr_67.gif] and [Graphics:Images/LaplaceDiffIntegrateMod_gr_68.gif].  We can use the linearity property of the Laplace transform to obtain

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_69.gif].

If we let [Graphics:Images/LaplaceDiffIntegrateMod_gr_70.gif] and [Graphics:Images/LaplaceDiffIntegrateMod_gr_71.gif] and apply Theorem 12.13 and Corollary 12.1 then we have

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_72.gif].

this in turn can be rearranged to obtain the form

(12.30)            [Graphics:Images/LaplaceDiffIntegrateMod_gr_73.gif].

The Laplace transform [Graphics:Images/LaplaceDiffIntegrateMod_gr_74.gif] of the solution [Graphics:Images/LaplaceDiffIntegrateMod_gr_75.gif] is easily found to be

(12.31)            [Graphics:Images/LaplaceDiffIntegrateMod_gr_76.gif].

    For many physical problems involving mechanical systems and electrical circuits, the transform [Graphics:Images/LaplaceDiffIntegrateMod_gr_77.gif] is known, and the inverse of  [Graphics:Images/LaplaceDiffIntegrateMod_gr_78.gif] can easily be computed.  This process is referred to as operational calculus and has the advantage of changing problems in differential equations into problems in algebra.  Then the solution obtained will satisfy the specific initial conditions.

 

Example 12.15.  Solve the initial value problem  

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_79.gif]   

        

                   [Graphics:Images/LaplaceDiffIntegrateMod_gr_80.gif]

                                    A graph of the solution.

Solution.

The right side of the differential equation is [Graphics:Images/LaplaceDiffIntegrateMod_gr_81.gif], so we have [Graphics:Images/LaplaceDiffIntegrateMod_gr_82.gif].  The initial conditions yield  [Graphics:Images/LaplaceDiffIntegrateMod_gr_83.gif]  and Equation (12.30)

becomes  [Graphics:Images/LaplaceDiffIntegrateMod_gr_84.gif]  which simplifies, and we get  

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_85.gif].  

Solving we get  [Graphics:Images/LaplaceDiffIntegrateMod_gr_86.gif].  

We then solve [Graphics:Images/LaplaceDiffIntegrateMod_gr_87.gif] with the help of Table 12.2 to compute

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_88.gif]

Explore Solution 12.15.

 

Example 12.16.  Solve the initial value problem  

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_103.gif]   

        

                   [Graphics:Images/LaplaceDiffIntegrateMod_gr_104.gif]

                                      A graph of the solution.

Solution.

As in Example 12.15, the right side of the differential equation is [Graphics:Images/LaplaceDiffIntegrateMod_gr_105.gif], so we have [Graphics:Images/LaplaceDiffIntegrateMod_gr_106.gif].  The initial conditions yield  [Graphics:Images/LaplaceDiffIntegrateMod_gr_107.gif]  and [Graphics:Images/LaplaceDiffIntegrateMod_gr_108.gif],  and Equation (12.30) becomes  [Graphics:Images/LaplaceDiffIntegrateMod_gr_109.gif],  which can be rewritten as  [Graphics:Images/LaplaceDiffIntegrateMod_gr_110.gif]  which simplifies, and we get  [Graphics:Images/LaplaceDiffIntegrateMod_gr_111.gif].

This time we use Equation (12.31) and obtain  [Graphics:Images/LaplaceDiffIntegrateMod_gr_112.gif], which simplifies, and we get  [Graphics:Images/LaplaceDiffIntegrateMod_gr_113.gif].  Now use the partial fraction expansion  [Graphics:Images/LaplaceDiffIntegrateMod_gr_114.gif]  to get the solution

            [Graphics:Images/LaplaceDiffIntegrateMod_gr_115.gif]  

Explore Solution 12.16.

 

Exercises for Section 12.6.  Laplace Transforms of Derivatives and Integrals

 

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Laplace Transform

 

 

 

  

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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2012 John H. Mathews, Russell W. Howell