7.3.1 translation on the -axis - pennsylvania state...

290 ● CHAPTER 7 THE LAPLACE TRANSFORM

7.3.1 TRANSLATION ON THE s -AXIS

A Translation Evaluating transforms such as and isstraightforward provided that we know (and we do) and . Ingeneral, if we know the Laplace transform of a function f, , it is pos-sible to compute the Laplace transform of an exponential multiple of f , that is,

with no additional effort other than translating, or shifting, the transformF(s) to This result is known as the first translation theor em or firsshifting theorem.

F(s a).�{eat f (t)},

�{ f (t)} � F(s)�{cos 4t}�{t3}

�{e2t cos 4t}�{e5tt3}

PROOF The proof is immediate, since by Definition 7.1.1

.

If we consider s a real variable, then the graph of F(s a) is the graph of F(s)shifted on the s-axis by the amount �a �. If a 0, the graph of F(s) is shifted a units tothe right, whereas if a � 0, the graph is shifted �a � units to the left. See Figure 7.3.1.

For emphasis it is sometimes useful to use the symbolism

,

where means that in the Laplace transform F(s) of f (t) we replace thesymbol s wherever it appears by s a.

s : s a

�{eat f (t)} � �{ f (t)}� s:sa

�{eat f (t)} � ��

0esteat

f (t) dt � ��

0e(sa)t

f (t) dt � F(s a)

s

F ( s )

s = a , a > 0

F

F( s − a)

FIGURE 7.3.1 Shift on s-axis

THEOREM 7.3.1 First Translation Theorem

If and a is any real number, then

.�{eat f (t)} � F(s a)

�{ f (t)} � F(s)

EXAMPLE 1 Using the First Translation Theorem

Evaluate (a) (b) .

SOLUTION The results follow from Theorems 7.1.1 and 7.3.1.

(a)

(b)

Inverse Form of Theorem 7.3.1 To compute the inverse of F(s a), wemust recognize F(s), find f (t) by taking the inverse Laplace transform of F(s), andthen multiply f (t) by the exponential function eat. This procedure can be summarizedsymbolically in the following manner:

, (1)

where .The first part of the next example illustrates partial fraction decomposition in the

case when the denominator of Y(s) contains repeated linear factors.

f (t) � � 1{F(s)}

� 1{F(s a)} � � 1{F(s) �s:sa} � eat f (t)

�{e2t cos 4t} � �{cos 4t}� s:s(2) �s

s2 � 16 �s:s�2

�s � 2

(s � 2)2 � 16

�{e5tt3} � �{t3}� s: s5 �3!s4 �

s:s5�

6(s 5)4

�{e2t cos 4t}�{e5tt3}

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7.3 OPERATIONAL PROPERTIES I ● 291

EXAMPLE 2 Partial Fractions: Repeated Linear Factors

Evaluate (a) (b) .

SOLUTION (a) A repeated linear factor is a term (s a)n, where a is a real numberand n is a positive integer � 2. Recall that if (s a)n appears in the denominator of arational expression, then the assumed decomposition contains n partial fractionswith constant numerators and denominators s a, (s a)2, . . . , (s a)n. Hence witha � 3 and n � 2 we write

.

By putting the two terms on the right-hand side over a common denominator, weobtain the numerator 2s � 5 � A(s 3) � B, and this identity yields A � 2 andB � 11. Therefore

(2)

and (3)

Now 1�(s 3)2 is F(s) � 1�s2 shifted three units to the right. Since ,it follows from (1) that

.

Finally, (3) is . (4)

(b) To start, observe that the quadratic polynomial s2 � 4s � 6 has no real zeros andso has no real linear factors. In this situation we complete the square:

. (5)

Our goal here is to recognize the expression on the right-hand side as some Laplacetransform F(s) in which s has been replaced throughout by s � 2. What we are trying to do is analogous to working part (b) of Example 1 backwards. The denom-inator in (5) is already in the correct form—that is, s2 � 2 with s replaced by s � 2.However, we must fix up the numerator by manipulating the constants:

.Now by termwise division, the linearity of �1, parts (e) and (d) of Theorem 7.2.1,

and finally (1),

(6)

(7) �12 e2t cos 12 t �

123

e2t sin 12t.

�12 � 1� s

s2 � 2 �s:s�2� �2

312 � 1� 12

s2 � 2 �s:s�2�

� 1� s>2 � 5> 3s2 � 4s � 6� �

12 � 1� s � 2

(s � 2)2 � 2� �23 � 1� 1

(s � 2)2 � 2�

s>2 � 5> 3

(s � 2)2 � 2�

12 (s � 2) � 2

3

(s � 2)2 � 2�

12

s � 2(s � 2)2 � 2

�23

1(s � 2)2 � 2

12s � 5

3 � 12

(s � 2) � 53 2

2 � 12

(s � 2) � 23

s>2 � 5>3s2 � 4s � 6

�s>2 � 5>3

(s � 2)2 � 2

� 1� 2s � 5(s 3)2� � 2e3t � 11e3tt

� 1� 1(s 3)2� � � 1�1

s2 �s:s3� � e3tt

� 1{1>s2} � t

� 1� 2s � 5(s 3)2� � 2� 1� 1

s 3� � 11� 1� 1(s 3)2�.

2s � 5(s 3)2 �

2s 3

�11

(s 3)2

2s � 5(s 3)2 �

A

s 3�

B

(s 3)2

� 1� s>2 � 5>3s2 � 4s � 6�� 1� 2s � 5

(s 3)2�



EXAMPLE 3 An Initial-Value Problem

Solve y� 6y� � 9y � t2e3t, y(0) � 2, y�(0) � 17.

SOLUTION Before transforming the DE, note that its right-hand side is similar tothe function in part (a) of Example 1. After using linearity, Theorem 7.3.1, and theinitial conditions, we simplify and then solve for :

.

The first term on the right-hand side was already decomposed into individual partiafractions in (2) in part (a) of Example 2:

.

Thus . (8)

From the inverse form (1) of Theorem 7.3.1, the last two terms in (8) are

.

Thus (8) is y(t) � 2e3t � 11te3t � 112t4e3t.

� 1�1s2 �

s:s3� � te3t and � 1�4!s5 �

s:s3� � t 4e3t

y(t) � 2� 1� 1s 3� � 11� 1� 1

(s 3)2� �24!

� 1� 4!(s 3)5�

Y(s) �2

s 3�

11(s 3)2 �

2(s 3)5

Y(s) �2s � 5(s 3)2 �

2(s 3)5

(s 3)2Y(s) � 2s � 5 �2

(s 3)3

(s2 6s � 9)Y(s) � 2s � 5 �2

(s 3)3

s2Y(s) sy(0) y�(0) 6[sY(s) y (0)] � 9Y(s) �2

(s 3)3

�{y �} 6�{y�} � 9�{y} � �{t2e3t}

Y(s) � �{ f (t)}


Solve y� � 4y� � 6y � 1 � et, y(0) � 0, y�(0) � 0.

SOLUTION

Since the quadratic term in the denominator does not factor into real linear factors, thepartial fraction decomposition for Y(s) is found to be

.

Moreover, in preparation for taking the inverse transform we already manipulatedthe last term into the necessary form in part (b) of Example 2. So in view of theresults in (6) and (7) we have the solution

Y(s) �1>6

s�

1>3s � 1

s> 2 � 5> 3

s2 � 4s � 6

Y(s) �2s � 1

s(s � 1)(s2 � 4s � 6)

(s2 � 4s � 6)Y(s) �2s � 1

s(s � 1)

s2Y(s) sy(0) y�(0) � 4[sY(s) y (0)] � 6Y(s) �1s

�1

s � 1

�{y�} � 4�{y�} � 6�{y} � �{1} � �{et}



. �16

�13

et 12

e2t cos 12t 123

e2t sin 12t

y(t) �16

� 1�1s� �

13

� 1� 1s � 1�

12 � 1� s � 2

(s � 2)2 � 2� 2

312 � 1� 12

(s � 2)2 � 2�

7.3.2 TRANSLATION ON THE t-AXIS

Unit Step Function In engineering, one frequently encounters functions that are either “off ” or “on.” For example, an external force acting on a mechanicalsystem or a voltage impressed on a circuit can be turned off after a period of time.It is convenient, then, to define a special function that is the number 0 (off ) up to acertain time t � a and then the number 1 (on) after that time. This function is calledthe unit step function or the Heaviside function, named after the English polymathOliver Heaviside (1850–1925).

Notice that we define only on the nonnegative t-axis, since this is allthat we are concerned with in the study of the Laplace transform. In a broader sense

for t � a. The graph of is given in Figure 7.3.2. In the casewhen a � 0, we take

When a function f defined for t � 0 is multiplied by , the unit stepfunction “turns off ” a portion of the graph of that function. For example, considerthe function f (t) � 2t 3. To “turn off ” the portion of the graph of f for 0 � t � 1,we simply form the product (2t 3) . See Figure 7.3.3. In general, thegraph of f (t) is 0 (off ) for 0 � t � a and is the portion of the graph of f (on)for t � a.

The unit step function can also be used to write piecewise-defined functions ina compact form. For example, if we consider 0 � t � 2, 2 � t � 3, and t � 3and the corresponding values of and , it should be apparentthat the piecewise-defined function shown in Figure 7.3.4 is the same as

. Also, a general piecewise-defined function ofthe type

(9)

is the same as

. (10)

Similarly, a function of the type

(11)

can be written

(12)f (t) � g(t)[�(t a) �(t b)].

f (t) � �0,g(t),0,

0 � t � a

a � t � b

t � b

f (t) � g(t) g(t) �(t a) � h(t) �(t a)

f (t) � �g(t),h(t),

0 � t � a

t � a

f (t) � 2 3�(t 2) � �(t 3)

�(t 3)�(t 2)

�(t a)�(t 1)

�(t a)�(t) � 1 for t � 0.

�(t a)�(t a) � 0

�(t a)

FIGURE 7.3.2 Graph of unit stepfunction

t

1

a

FIGURE 7.3.3 Function isf (t) � (2t 3) � (t 1)

1

y

t

FIGURE 7.3.4 Function isf (t) � 2 3�(t 2) � �(t 3)

−1

2

t

f(t)

DEFINITION 7.3.1 Unit Step Function

The unit step function is defined to b

�(t a) � �0,1, 0 � t � a

t � a.

�(t a)



FIGURE 7.3.5 Function f in Example 5

100

5

f (t)

t

FIGURE 7.3.6 Shift on t-axis

(a) f(t), t � 0

(b) f(t � a) (t � a)

t

f(t)

t

f(t)

a

zero for0 � t � a

one fort � a

�{ f (t � a) (t � a)} � � e�stf (t � a) (t � a) dt � � e�stf (t � a) (t � a) dt � � e�stf (t � a) dt.� � �a

0

�

a

�

a

Now if we let v � t a, dv � dt in the last integral, then

PROOF By the additive interval property of integrals,

can be written as two integrals:

�(t a) dt��

0 est f (t a)

.�{ f (t a) �(t a)} � ��

0es(v�a) f (v) dv � eas��

0esv f (v) dv � eas�{ f (t)}

We often wish to find the Laplace transform of just a unit step function. This can befrom either Definition 7.1.1 or Theorem 7.3.2. If we identify f (t) � 1 in Theorem 7.3.2,then f (t a) � 1, , and so

. (14)�{�(t a)} �eas

s

F(s) � �{1} � 1>s

EXAMPLE 5 A Piecewise-Defined Functio

Express in terms of unit step functions. Graph.

SOLUTION The graph of f is given in Figure 7.3.5. Now from (9) and (10) with a � 5, g(t) � 20t, and h(t) � 0 we get .

Consider a general function y � f (t) defined for t � 0. The piecewise-definefunction

(13)

plays a significant role in the discussion that follows. As shown in Figure 7.3.6, fora 0 the graph of the function coincides with the graph ofy for (which is the entire graph of shifted a units tothe right on the t-axis), but is identically zero for

We saw in Theorem 7.3.1 that an exponential multiple of f (t) results in a transla-tion of the transform F(s) on the s-axis. As a consequence of the next theorem we seethat whenever F(s) is multiplied by an exponential function eas, a 0, the inversetransform of the product eas F(s) is the function f shifted along the t-axis in the man-ner illustrated in Figure 7.3.6(b). This result, presented next in its direct transformversion, is called the second translation theorem or second shifting theorem.

0 � t � a.y � f(t), t � 0t � a� f (t a)

y � f (t a) �(t a)

f (t a) �(t a) � �0,f (t a),

0 � t � a

t � a

f (t) � 20t 20t �(t 5)

f (t) � �20t,0,

0 � t � 5 t � 5

THEOREM 7.3.2 Second Translation Theorem

If and a 0, then

.�{ f (t a) �(t a)} � easF(s)

F(s) � �{ f (t)}

EXAMPLE 6 Figure 7.3.4 Revisited

Find the Laplace transform the function f in Figure 7.3.4.

SOLUTION We use f expressed in terms of the unit step function

f(t) � 2 3� (t 2) � � (t 3)



and the result given in (14):

Inverse Form of Theorem 7.3.2 If , the inverse form off (t) � � 1{F(s)}

�2s

3e2s

s�

e3s

s.

�{ f (t)} � 2�{1} 3�{�(t 2)} � �{�(t 3)}

EXAMPLE 7 Using Formula (15)

Evaluate .

SOLUTION (a) With the three identifications a � 2, F(s) � 1�(s 4), and�1{F(s)} � e4t, we have from (15)

.

(b) With a � p�2, F(s) � s�(s2 � 9), and , (15) yields

.

The last expression can be simplified somewhat by using the addition formula for the

cosine. Verify that the result is the same as

Alternative Form of Theorem 7.3.2 We are frequently confronted withthe problem of finding the Laplace transform of a product of a function g and a unitstep function where the function g lacks the precise shifted form in Theorem 7.3.2. To find the Laplace transform of , it is possible to fiup g(t) into the required form by algebraic manipulations. For example, ifwe wanted to use Theorem 7.3.2 to find the Laplace transform of , wewould have to force into the form You should work through thedetails and verify that is an identity. Therefore

where each term on the right-hand side can now be evaluated by Theorem 7.3.2. Butsince these manipulations are time consuming and often not obvious, it is simpler todevise an alternative version of Theorem 7.3.2. Using Definition 7.1.1, the definitioof , and the substitution u � t a, we obtain

.

That is, . (16)�{g(t)�(t a)} � eas �{g(t � a)}

�{g(t) �(t a)} � ��

aest

g(t) dt � ��

0es(u�a)

g(u � a) du

�(t a)

�{t2�(t 2)} � �{(t 2)2 � (t 2) � 4(t 2) � (t 2) � 4�(t 2)},

t2 � (t 2)2 � 4(t 2) � 4f (t 2).g(t) � t2

t2�(t 2)f (t a)

g(t)�(t a)f (t a)�(t a)

sin 3t ��t

2�.

� 1� s

s2 � 9 e s/2� � cos 3�t

2� ��t

2�� 1{F(s)} � cos 3t

� 1� 1s 4

e2s� � e4(t2) �(t 2)

(b) � 1� s

s2 � 9 e s/2�(a) � 1� 1

s 4 e2s�

EXAMPLE 8 Second Translation Theorem—Alternative Form

Evaluate .

SOLUTION With g(t) � cos t and a � p, then g(t � p) � cos(t � p) � cos tby the addition formula for the cosine function. Hence by (16),

�{cos t �(t )} � e s �{cos t} �

s

s2 � 1 e s.

�{cos t �(t )}

Theorem 7.3.2, a 0, is

. (15)� 1{easF(s)} � f (t a) �(t a)



FIGURE 7.3.7 Graph of function (18)in Example 9

_2

12345

_ 1t

y

2π π 3π(18)

� �5et,

5et �32

e(t ) �32

sin t �32

cos t,

0 � t �

t � .

; trigonometric identities � 5et �32

[e(t ) � sin t � cos t] �(t )

y(t) � 5et �32

e(t ) �(t ) 32

sin(t ) �(t ) 32

cos(t ) �(t )

We obtained the graph of (18) shown in Figure 7.3.7 by using a graphing utility.

Beams In Section 5.2 we saw that the static deflection y(x) of a uniform beamof length L carrying load w(x) per unit length is found from the linear fourth-orderdifferential equation

(19)

where E is Young’s modulus of elasticity and I is a moment of inertia of a cross sectionof the beam. The Laplace transform is particularly useful in solving (19) when w(x)is piecewise-defined. However, to use the Laplace transform, we must tacitly assumethat y(x) and w(x) are defined on (0, �) rather than on (0, L). Note, too, that the nextexample is a boundary-value problem rather than an initial-value problem.

EI d4y

dx4 � w(x),

FIGURE 7.3.8 Embedded beam withvariable load in Example 10

wall

x

y

L

w(x)

�1� 1s � 1

e s� � e(t ) �(t ), �1� 1

s2 � 1 e s� � sin(t ) �(t ),

and .

Thus the inverse of (17) is

� 1� s

s2 � 1 e s� � cos(t ) �(t )


Solve

SOLUTION The function f can be written as , so by linear-ity, the results of Example 7, and the usual partial fractions, we have

. (17)

Now proceeding as we did in Example 7, it follows from (15) with a � p that theinverses of the terms inside the brackets are

Y(s) �5

s � 1

32

1s � 1

e s �1

s2 � 1 e s �

s

s2 � 1 e s

(s � 1)Y(s) � 5 3s

s2 � 1 e s

sY(s) y(0) � Y(s) � 3 s

s2 � 1 e s

�{y�} � �{y} � 3�{cos t �(t )}

f (t) � 3 cos t �(t )

y� � y � f (t), y(0) � 5, where f (t) � �0,3 cos t,

0 � t �

t � .

EXAMPLE 10 A Boundary-Value Problem

A beam of length L is embedded at both ends, as shown in Figure 7.3.8. Find thedeflection of the beam when the load is given b

w(x) � �w0�1 2L

x�,

0,

0 � x � L>2

L>2 � x � L.



SOLUTION Recall that because the beam is embedded at both ends, the boundaryconditions are y(0) � 0, y�(0) � 0, y(L) � 0, y�(L) � 0. Now by (10) we can expressw(x) in terms of the unit step function:

Transforming (19) with respect to the variable x gives

or

If we let c1 � y�(0) and c2 � y�(0), then

,

and consequently

Y(s) �c1

s3 �c2

s4 �2w0

EIL L>2

s5 1s6 �

1s6

eLs/2

s4Y(s) sy �(0) y�(0) �2w0

EIL L>2

s

1s2 �

1s2

eLs/2 .

EI�s4Y(s) s3y(0) s2y�(0) sy �(0) y� (0)� �2w0

L L>2

s

1s2 �

1s2

eLs/2

�2w0

L L

2 x � �x

L

2� ��x L

2� .

w(x) � w0�1 2L

x� w0�1 2L

x� ��x L

2�

�c1

2 x2 �

c2

6 x3 �

w0

60 EIL 5L

2 x4 x5 � �x

L

2�5��x

L

2� .

y(x) �c1

2! � 1�2!

s3� �c2

3! � 1�3!

s4� �2w0

EIL L>2

4! � 1�4!

s5� 15!

� 1�5!s6� �

15!

� 1�5!s6

eLs/ 2�

Applying the conditions y(L) � 0 and y�(L) � 0 to the last result yields a system ofequations for c1 and c2:

Solving, we find c1 � 23w0L2�(960EI) and c2 � 9w0L�(40EI). Thus the deflection is given by

y(x) �23w0L2

1920EI x2

3w0L

80EI x3 �

w0

60EIL 5L

2 x4 x5 � �x

L

2�5��x

L

2� .

c1 L � c2 L2

2�

85w0L3

960EI� 0.

c1 L2

2� c2

L3

6�

49w0L4

1920EI� 0

EXERCISES 7.3 Answers to selected odd-numbered problems begin on page ANS-12.

7.3.1 TRANSLATION ON THE s -AXIS

In Problems 1–20 find either F(s) or f (t), as indicated.

1. 2.

3. 4.

5. 6.

7. 8.

9.

10. ��e3t �9 4t � 10 sin t

2��{(1 et � 3e4t) cos 5t}

�{e2t cos 4t}�{et sin 3t}

�{e2t(t 1)2}�{t(et � e2t)2}

�{t10e7t}�{t3e2t}

�{te6t}�{te10t}

11. 12.

13. 14.

15. 16.

17. 18.

19. 20. � 1�(s � 1)2

(s � 2)4�� 1� 2s 1s2(s � 1)3�

� 1� 5s

(s 2)2�� 1� s

(s � 1)2�

� 1� 2s � 5s2 � 6s � 34�� 1� s

s2 � 4s � 5�

� 1� 1s2 � 2s � 5�� 1� 1

s2 6s � 10�

� 1� 1(s 1)4�� 1� 1

(s � 2)3�



In Problems 21–30 use the Laplace transform to solve thegiven initial-value problem.

21. y� � 4y � e4t, y(0) � 222. y� y � 1 � tet, y(0) � 023. y� � 2y� � y � 0, y(0) � 1, y�(0) � 124. y� 4y� � 4y � t3e2t, y(0) � 0, y�(0) � 025. y� 6y� � 9y � t, y(0) � 0, y�(0) � 126. y� 4y� � 4y � t3, y(0) � 1, y�(0) � 027. y� 6y� � 13y � 0, y(0) � 0, y�(0) � 328. 2y� � 20y� � 51y � 0, y(0) � 2, y�(0) � 029. y� y� � et cos t, y(0) � 0, y�(0) � 030. y� 2y� � 5y � 1 � t, y(0) � 0, y�(0) � 4

In Problems 31 and 32 use the Laplace transform andthe procedure outlined in Example 10 to solve the givenboundary-value problem.

31. y� � 2y� � y � 0, y�(0) � 2, y(1) � 2

32. y� � 8y� � 20y � 0, y(0) � 0, y�(p) � 0

33. A 4-pound weight stretches a spring 2 feet. The weightis released from rest 18 inches above the equilibriumposition, and the resulting motion takes place in amedium offering a damping force numerically equal to

times the instantaneous velocity. Use the Laplacetransform to find the equation of motion x(t).

34. Recall that the differential equation for the instanta-neous charge q(t) on the capacitor in an LRC-seriescircuit is given by

. (20)

See Section 5.1. Use the Laplace transform to find q(t)when L � 1 h, R � 20 �, C � 0.005 f, E(t) � 150 V,t 0, q(0) � 0, and i(0) � 0. What is the current i(t)?

35. Consider a battery of constant voltage E0 that chargesthe capacitor shown in Figure 7.3.9. Divide equa-tion (20) by L and define 2l � R�L and v2 � 1�LC.Use the Laplace transform to show that the solutionq(t) of q� � 2lq� � v2q � E0 �L subject to q(0) � 0,i(0) � 0 is

q(t) � �E0C1 e�t (cosh 1�2 �2t

��

1�2 �2 sinh 1�2 �2t) , � �,

E0C[1 e�t (1 � �t)], � � �,

E0C1 e�t (cos 1�2 �2t

�

�

1�2 �2

sin 1�2 �2t) , � � �.

L d 2q

dt2 � R dq

dt�

1C

q � E(t)

78

36. Use the Laplace transform to find the charge q(t)in an RC series circuit when q(0) � 0 andE(t) � E0ekt, k 0. Consider two cases: k � 1�RCand k � 1�RC.

7.3.2 TRANSLATION ON THE t-AXIS

In Problems 37–48 find either F(s) or f (t), as indicated.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

In Problems 49 – 54 match the given graph with one of the functions in (a)–(f ). The graph of f (t) is given inFigure 7.3.10.

(a)

(b)

(c)

(d)

(e)

(f) f (t a) �(t a) f (t a) �(t b)f (t) �(t a) f (t) �(t b)f (t) f (t) �(t b)f (t) �(t a)f (t b) �(t b)f (t) f (t) �(t a)

� 1� e2s

s2(s 1)�� 1� es

s(s � 1)�

�1�se s/2

s2 � 4�� 1� e s

s2 � 1�

� 1�(1 � e2s)2

s � 2 �� 1�e2s

s3 �

��sin t ��t

2��{cos 2t �(t )}

�{(3t � 1)�(t 1)}�{t �(t 2)}

�{e2t �(t 2)}�{(t 1)�(t 1)}

FIGURE 7.3.9 Series circuit in Problem 35

E0 R

C

L

FIGURE 7.3.10 Graph for Problems 49–54

t

f (t)

a b

49.

FIGURE 7.3.11 Graph for Problem 49

t

f (t)

a b




t

f (t)

a b


t

f (t)

a b


t

f (t)

a b


t

f (t)

a b


t

f (t)

a b

50.

51.

52.

53.

54.

In Problems 55–62 write each function in terms of unit stepfunctions. Find the Laplace transform of the given function.

55.

56.

57. f (t) � �0,t2, 0 � t � 1 t � 1

f (t) � �1,0,1,

0 � t � 4 4 � t � 5 t � 5

f (t) � �2,2,

0 � t � 3 t � 3

58.

59.

60. f (t) � �sin t,0,

0 � t � 2

t � 2

f (t) � � t,0,

0 � t � 2 t � 2

f (t) � �0,sin t,

0 � t � 3 >2 t � 3 >2

61.

62.


3

2

1

staircase function

t

f(t)

1 2 3 4


1

rectangular pulse

tba

f(t)

In Problems 63–70 use the Laplace transform to solve thegiven initial-value problem.

63. y� � y � f (t), y(0) � 0, where f (t) �

64. y� � y � f (t), y(0) � 0, where

65. y� � 2y � f (t), y(0) � 0, where

66. where

67. , y(0) � 1, y�(0) � 0

68. , y(0) � 0, y�(0) � 1

69. where

70. y� � 4y� � 3y � 1 �(t 2) �(t 4) � �(t 6),y(0) � 0, y�(0) � 0

f (t) � �0,1,0,

0 � t �

� t � 2

t � 2

y� � y � f (t), y(0) � 0, y�(0) � 1,

y� 5y� � 6y � �(t 1)

y � � 4y � sin t �(t 2 )

f (t) � �1,0,

0 � t � 1 t � 1

y � � 4y � f (t), y(0) � 0, y�(0) � 1,

f (t) � � t,0,

0 � t � 1 t � 1

f (t) � � 1,1,

0 � t � 1 t � 1

�0,5,

0 � t � 1 t � 1



71. Suppose a 32-pound weight stretches a spring 2 feet. If theweight is released from rest at the equilibrium position,find the equation of motion x(t) if an impressed forcef (t) � 20t acts on the system for 0 � t � 5 and is thenremoved (see Example 5). Ignore any damping forces.Use a graphing utility to graph x(t) on the interval [0, 10].

72. Solve Problem 71 if the impressed force f (t) � sin t actson the system for 0 � t � 2p and is then removed.

In Problems 73 and 74 use the Laplace transform to find thecharge q(t) on the capacitor in an RC-series circuit subject tothe given conditions.

73. q(0) � 0, R � 2.5 �, C � 0.08 f, E(t) given inFigure 7.3.19

t

E(t)

3

5

FIGURE 7.3.19 E(t) in Problem 73

t

E(t)

1.5

30

30et


74. q(0) � q0, R � 10 �, C � 0.1 f, E(t) given inFigure 7.3.20

75. (a) Use the Laplace transform to find the currenti(t) in a single-loop LR-series circuit wheni(0) � 0, L � 1 h, R � 10 �, and E(t) is as givenin Figure 7.3.21.

(b) Use a computer graphing program to graph i(t) for0 � t � 6. Use the graph to estimate imax andimin, the maximum and minimum values of thecurrent.



/2

1

−1t

E(t)

3 /2π

sin t, 0 ≤ t < 3 /2

π π

π

t31

E(t)

E0

76. (a) Use the Laplace transform to find the charge q(t) on the capacitor in an RC-series circuit when q(0) � 0, R � 50 �, C � 0.01 f, and E(t) is asgiven in Figure 7.3.22.

(b) Assume that E0 � 100 V. Use a computer graphingprogram to graph q(t) for 0 � t � 6. Use thegraph to estimate qmax, the maximum value ofthe charge.

77. A cantilever beam is embedded at its left end and freeat its right end. Use the Laplace transform to find thedeflection y(x) when the load is given by

78. Solve Problem 77 when the load is given by

79. Find the deflection y(x) of a cantilever beam embeddedat its left end and free at its right end when the load is asgiven in Example 10.

80. A beam is embedded at its left end and simply supportedat its right end. Find the deflection y(x) when the load isas given in Problem 77.

Mathematical Model

81. Cake Inside an Oven Reread Example 4 in Sec-tion 3.1 on the cooling of a cake that is taken out of anoven.(a) Devise a mathematical model for the temperature of

a cake while it is inside the oven based on the fol-lowing assumptions: At t � 0 the cake mixture is atthe room temperature of 70°; the oven is not pre-heated, so at t � 0, when the cake mixture is placedinto the oven, the temperature inside the oven is also70°; the temperature of the oven increases linearlyuntil t � 4 minutes, when the desired temperatureof 300° is attained; the oven temperature is a con-stant 300° for t � 4.

(b) Use the Laplace transform to solve the initial-valueproblem in part (a).

w(x) � �0, w0,0,

0 � x � L>3 L>3 � x � 2L>3 2L>3 � x � L.

w(x) � �w0,0,

0 � x � L> 2 L>2 � x � L.


7.4 OPERATIONAL PROPERTIES II ● 301

Discussion Problems

82. Discuss how you would fix up each of the followingfunctions so that Theorem 7.3.2 could be used directlyto find the given Laplace transform. Check youranswers using (16) of this section.

(a) (b)

(c) (d)

83. (a) Assume that Theorem 7.3.1 holds when the sym-bol a is replaced by ki, where k is a real number

�{(t2 3t)�(t 2)}�{cos t �(t )}

�{et �(t 5)}�{(2t � 1)�(t 1)}

and i2 � 1. Show that can be used todeduce

(b) Now use the Laplace transform to solve the initial-value problem x� � v2x � cos vt, x(0) � 0,x�(0) � 0.

�{t sin kt} �2ks

(s2 � k2)2.

�{t cos kt} �s2 k2

(s2 � k2)2

�{tekti}

7.4.1 DERIVATIVES OF A TRANSFORM

Multiplying a Function by tn The Laplace transform of the product of afunction f (t) with t can be found by differentiating the Laplace transform of f (t). Tomotivate this result, let us assume that exists and that it is possibleto interchange the order of differentiation and integration. Then

;

that is, .

We can use the last result to find the Laplace transform of t2 f (t):

�{t f (t)} � d

ds �{ f (t)}

d

ds F(s) �

d

ds ��

0 est

f (t) dt � ��

0 �

�s [est f (t)] dt � ��

0 est t f (t) dt � �{t f (t)}

F(s) � �{ f (t)}

.�{t2 f (t)} � �{t � t f (t)} �

d

ds �{t f (t)} �

d

ds �

d

ds �{ f (t)}� �

d 2

ds2 �{ f (t)}

The preceding two cases suggest the general result for .�{tn f (t)}

OPERATIONAL PROPERTIES II

REVIEW MATERIAL● Definition 7.1.● Theorems 7.3.1 and 7.3.2

INTRODUCTION In this section we develop several more operational properties of theLaplace transform. Specificall , we shall see how to find the transform of a function f (t) that ismultiplied by a monomial t n, the transform of a special type of integral, and the transform of a pe-riodic function. The last two transform properties allow us to solve some equations that we havenot encountered up to this point: Volterra integral equations, integrodifferential equations, and or-dinary differential equations in which the input function is a periodic piecewise-defined function.

7.4

THEOREM 7.4.1 Derivatives of Transforms

If and n � 1, 2, 3, . . . , then

.�{tn f (t)} � (1)n dn

dsn F(s)

F(s) � �{ f (t)}


7.3.1 translation on the -axis - pennsylvania state...

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