5.Laplace Casos Especiais.1234.Sol

4.3 Translation Theorems

45. For y 4y = 6e3t 3et the transfer function is W (s) = 1/(s2 4s). The zero-input response is

y0(t) ={

s 5s2 4s

}=

{54 1s 1

4 1s 4

}=

54 1

4e4t ,

and the zero-state response is

y1(t) ={

6(s 3)(s2 4s)

3(s + 1)(s2 4s)

}

={

2720 1s 4

2s 3 +

54 1s 3

5 1s + 1

}

=2720

e4t 2e3t + 54 3

5et .

46. From Theorem 4.4, if f and f are continuous and of exponential order, {f (t)} = sF (s) f(0). FromTheorem 4.5, lims {f (t)} = 0 so

lims[sF (s) f(0)] = 0 and limsF (s) = f(0).

For f(t) = cos kt,

lims sF (s) = lims s

s

s2 + k2= 1 = f(0).

EXERCISES 4.3Translation Theorems

1.{te10t

}=

1(s 10)2 2.

{te6t

}=

1(s + 6)2

3.{t3e2t

}=

3!(s + 2)4

4.{t10e7t

}=

10!(s + 7)11

5.{t(et + e2t

)2}=

{te2t + 2te3t + te4t

}=

1(s 2)2 +

2(s 3)2 +

1(s 4)2

6.{e2t(t 1)2} = {t2e2t 2te2t + e2t} = 2

(s 2)3 2

(s 2)2 +1

s 2

7.{et sin 3t

}=

3(s 1)2 + 9 8.

{e2t cos 4t

}=

s + 2(s + 2)2 + 16

9. {(1 et + 3e4t) cos 5t} = {cos 5t et cos 5t + 3e4t cos 5t} = ss2 + 25

s 1(s 1)2 + 25 +

3(s + 4)(s + 4)2 + 25

10.{e3t(9 4t + 10 sin t

2

)}=

{9e3t 4te3t + 10e3t sin t

2

}=

9s 3

4(s 3)2 +

5(s 3)2 + 1/4

11.{

1(s + 2)3

}=

{12

2(s + 2)3

}=

12t2e2t

207


12.{

1(s 1)4

}=

16

{3!

(s 1)4}

=16t3et

13.{

1s2 6s + 10

}=

{1

(s 3)2 + 12}

= e3t sin t

14.{

1s2 + 2s + 5

}=

{12

2(s + 1)2 + 22

}=

12et sin 2t

15.{

s

s2 + 4s + 5

}=

{s + 2

(s + 2)2 + 12 2 1

(s + 2)2 + 12

}= e2t cos t 2e2t sin t

16.{

2s + 5s2 + 6s + 34

}=

{2

(s + 3)(s + 3)2 + 52

15

5(s + 3)2 + 52

}= 2e3t cos 5t 1

5e3t sin 5t

17.{

s

(s + 1)2

}=

{s + 1 1(s + 1)2

}=

{1

s + 1 1

(s + 1)2

}= et tet

18.{

5s(s 2)2

}=

{5(s 2) + 10

(s 2)2}

={

5s 2 +

10(s 2)2

}= 5e2t + 10te2t

19.{

2s 1s2(s + 1)3

}=

{5s 1

s2 5

s + 1 4

(s + 1)2 3

22

(s + 1)3

}= 5 t 5et 4tet 3

2t2et

20.{

(s + 1)2

(s + 2)4

}=

{1

(s + 2)2 2

(s + 2)3+

16

3!(s + 2)4

}= te2t t2e2t + 1

6t3e2t

21. The Laplace transform of the dierential equation is

s {y} y(0) + 4 {y} = 1s + 4

.

Solving for {y} we obtain{y} = 1

(s + 4)2+

2s + 4

.

Thusy = te4t + 2e4t.


s {y} {y} = 1s+

1(s 1)2 .

Solving for {y} we obtain

{y} = 1s(s 1) +

1(s 1)3 =

1s+

1s 1 +

1(s 1)3 .

Thusy = 1 + et + 1

2t2et.


s2 {y} sy(0) y(0) + 2[s {y} y(0)]+ {y} = 0.Solving for {y} we obtain

{y} = s + 3(s + 1)2

=1

s + 1+

2(s + 1)2

.

Thusy = et + 2tet.

208



s2 {y} sy(0) y(0) 4 [s {y} y(0)] + 4 {y} = 6(s 2)4 .

Solving for {y} we obtain {y} = 120

5!(s 2)6 . Thus, y =

120

t5e2t.


s2 {y} sy(0) y(0) 6 [s {y} y(0)] + 9 {y} = 1s2

.


{y} = 1 + s2

s2(s 3)2 =227

1s+

19

1s2 2

271

s 3 +109

1(s 3)2 .

Thus

y =227

+19t 2

27e3t +

109

te3t.


s2 {y} sy(0) y(0) 4 [s {y} y(0)] + 4 {y} = 6s4

.


{y} = s5 4s4 + 6s4(s 2)2 =

34

1s+

98

1s2

+34

2s3

+14

3!s4

+14

1s 2

138

1(s 2)2 .

Thus

y =34+

98t +

34t2 +

14t3 +

14e2t 13

8te2t.


s2 {y} sy(0) y(0) 6 [s {y} y(0)] + 13 {y} = 0.Solving for {y} we obtain

{y} = 3s2 6s + 13 =

32

2(s 3)2 + 22 .

Thus

y = 32e3t sin 2t.


2[s2 {y} sy(0)]+ 20[s {y} y(0)]+ 51 {y} = 0.


{y} = 4s + 402s2 + 20s + 51

=2s + 20

(s + 5)2 + 1/2=

2(s + 5)(s + 5)2 + 1/2

+10

(s + 5)2 + 1/2.

Thus

y = 2e5t cos(t/2 ) + 10

2 e5t sin(t/

2 ).


s2 {y} sy(0) y(0) [s {y} y(0)] = s 1(s 1)2 + 1 .

209



{y} = 1s(s2 2s + 2) =

12

1s 1

2s 1

(s 1)2 + 1 +12

1(s 1)2 + 1 .

Thus

y =12 1

2et cos t +

12et sin t.


s2 {y} sy(0) y(0) 2 [s {y} y(0)] + 5 {y} = 1s+

1s2

.


{y} = 4s2 + s + 1

s2(s2 2s + 5) =725

1s+

15

1s2

+7s/25 + 109/25

s2 2s + 5

=725

1s+

15

1s2 7

25s 1

(s 1)2 + 22 +5125

2(s 1)2 + 22 .

Thus

y =725

+15t 7

25et cos 2t +

5125

et sin 2t.

31. Taking the Laplace transform of both sides of the dierential equation and letting c = y(0) we obtain

{y}+ {2y}+ {y} = 0s2 {y} sy(0) y(0) + 2s {y} 2y(0) + {y} = 0

s2 {y} cs 2 + 2s {y} 2c + {y} = 0(s2 + 2s + 1

) {y} = cs + 2c + 2{y} = cs

(s + 1)2+

2c + 2(s + 1)2

= cs + 1 1(s + 1)2

+2c + 2(s + 1)2

=c

s + 1+

c + 2(s + 1)2

.

Therefore,

y(t) = c{

1s + 1

}+ (c + 2)

{1

(s + 1)2

}= cet + (c + 2)tet.

To nd c we let y(1) = 2. Then 2 = ce1 + (c + 2)e1 = 2(c + 1)e1 and c = e 1. Thus

y(t) = (e 1)et + (e + 1)tet.

32. Taking the Laplace transform of both sides of the dierential equation and letting c = y(0) we obtain

{y}+ {8y}+ {20y} = 0s2 {y} y(0) + 8s {y}+ 20 {y} = 0

s2 {y} c + 8s {y}+ 20 {y} = 0(s2 + 8s + 20) {y} = c

{y} = cs2 + 8s + 20

=c

(s + 4)2 + 4.

210


Therefore,

y(t) ={

c

(s + 4)2 + 4

}=

c

2e4t sin 2t = c1e4t sin 2t.

To nd c we let y() = 0. Then 0 = y() = ce4 and c = 0. Thus, y(t) = 0. (Since the dierential equationis homogeneous and both boundary conditions are 0, we can see immediately that y(t) = 0 is a solution. Wehave shown that it is the only solution.)

33. Recall from Section 3.8 that mx = kxx. Now m = W/g = 4/32 = 18 slug, and 4 = 2k so that k = 2 lb/ft.Thus, the dierential equation is x + 7x + 16x = 0. The initial conditions are x(0) = 3/2 and x(0) = 0.The Laplace transform of the dierential equation is

s2 {x}+ 32s + 7s {x}+ 21

2+ 16 {x} = 0.

Solving for {x} we obtain

{x} = 3s/2 21/2s2 + 7s + 16

= 32

s + 7/2(s + 7/2)2 + (

15/2)2

715

10

15/2

(s + 7/2)2 + (15/2)2

.

Thus

x = 32e7t/2 cos

152

t 715

10e7t/2 sin

152

t.

34. The dierential equation isd2q

dt2+ 20

dq

dt+ 200q = 150, q(0) = q(0) = 0.

The Laplace transform of this equation is

s2 {q}+ 20s {q}+ 200 {q} = 150s

.

Solving for {q} we obtain

{q} = 150s(s2 + 20s + 200)

=34

1s 3

4s + 10

(s + 10)2 + 102 3

410

(s + 10)2 + 102.

Thus

q(t) =34 3

4e10t cos 10t 3

4e10t sin 10t

and

i(t) = q(t) = 15e10t sin 10t.

35. The dierential equation isd2q

dt2+ 2

dq

dt+ 2q =

E0L

, q(0) = q(0) = 0.


s2 {q}+ 2s {q}+ 2 {q} = E0L

1s

or (s2 + 2s + 2

) {q} = E0L

1s.

Solving for {q} and using partial fractions we obtain

{q} = E0L

(1/2

s (1/

2)s + 2/2

s2 + 2s + 2

)=

E0L2

(1s s + 2

s2 + 2s + 2

).

211


For > we write s2 + 2s + 2 = (s + )2 (2 2), so (recalling that 2 = 1/LC){q} = E0C

(1s s +

(s + )2 (2 2)

(s + )2 (2 2))

.

Thus for > ,

q(t) = E0C[1 et

(cosh

2 2 t

2 2 sinh

2 2 t)]

.

For < we write s2 + 2s + 2 = (s + )2 +(2 2), so

{q} = E0C(

1s s +

(s + )2 + (2 2)

(s + )2 + (2 2))

.

Thus for < ,

q(t) = E0C[1 et

(cos

2 2 t

2 2 sin

2 2 t)]

.

For = , s2 + 2 + 2 = (s + )2 and

{q} = E0L

1s(s + )2

=E0L

(1/2

s 1/

2

s + 1/

(s + )2

)=

E0L2

(1s 1

s +

(s + )2

).

Thus for = ,

q(t) = E0C(1 et tet) .

36. The dierential equation is

Rdq

dt+

1C

q = E0ekt, q(0) = 0.


Rs {q}+ 1C

{q} = E0 1s + k

.


{q} = E0C(s + k)(RCs + 1)

=E0/R

(s + k)(s + 1/RC).

When 1/RC = k we have by partial fractions

{q} = E0R

(1/(1/RC k)

s + k 1/(1/RC k)

s + 1/RC

)=

E0R

11/RC k

(1

s + k 1

s + 1/RC

).

Thus

q(t) =E0C

1 kRC(ekt et/RC

).

When 1/RC = k we have

{q} = E0R

1(s + k)2

.

Thus

q(t) =E0R

tekt =E0R

tet/RC .

37.{(t 1) (t 1)} = es

s2

38.{e2t (t 2)} = {e(t2) (t 2)} = e2s

s + 1

212


39.{t (t 2)} = {(t 2) (t 2) + 2 (t 2)} = e2s

s2+

2e2s

sAlternatively, (16) of this section could be used:

{t (t 2)} = e2s {t + 2} = e2s(

1s2

+2s

).

40.{(3t + 1) (t 1)} = 3 {(t 1) (t 1)}+ 4 { (t 1)} = 3es

s2+

4es

sAlternatively, (16) of this section could be used:

{(3t + 1) (t 1)} = es {3t + 4} = es(

3s2

+4s

).

41.{cos 2t (t )} = {cos 2(t ) (t )} = ses

s2 + 4Alternatively, (16) of this section could be used:

{cos 2t (t )} = es {cos 2(t + )} = es {cos 2t} = es ss2 + 4

.

42.{sin t

(t

2

)}=

{cos

(t

2

) (t

2

)}=

ses/2

s2 + 1

Alternatively, (16) of this section could be used:{sin t

(t

2

)}= es/2

{sin

(t +

2

)}= es/2 {cos t} = es/2 s

s2 + 1.

43.{

e2s

s3

}=

{12 2s3

e2s}

=12(t 2)2 (t 2)

44.{

(1 + e2s)2

s + 2

}=

{1

s + 2+

2e2s

s + 2+

e4s

s + 2

}= e2t + 2e2(t2) (t 2) + e2(t4) (t 4)

45.{

es

s2 + 1

}= sin(t ) (t ) = sin t (t )

46.{

ses/2

s2 + 4

}= cos 2

(t

2

) (t

2

)= cos 2t

(t

2

)

47.{

es

s(s + 1)

}=

{es

s e

s

s + 1

}= (t 1) e(t1) (t 1)

48.{

e2s

s2(s 1)}

={e

2s

s e

2s

s2+

e2s

s 1}

= (t 2) (t 2) (t 2) + et2 (t 2)

49. (c) 50. (e) 51. (f) 52. (b) 53. (a) 54. (d)

55.{2 4 (t 3)} = 2

s 4

se3s

56.{1 (t 4) + (t 5)} = 1

s e

4s

s+

e5s

s

57.{t2 (t 1)} = {[(t 1)2 + 2t 1] (t 1)} = {[(t 1)2 + 2(t 1) 1] (t 1)}

=(

2s3

+2s2

+1s

)es

213


Alternatively, by (16) of this section,

{t2 (t 1)} = es {t2 + 2t + 1} = es(

2s3

+2s2

+1s

).

58.{sin t

(t 3

2

)}=

{ cos

(t 3

2

) (t 3

2

)}= se

3s/2

s2 + 1

59.{t t (t 2)} = {t (t 2) (t 2) 2 (t 2)} = 1

s2 e

2s

s2 2e

2s

s

60.{sin t sin t (t 2)} = {sin t sin(t 2) (t 2)} = 1

s2 + 1 e

2s

s2 + 1

61.{f(t)

}=

{(t a) (t b)} = eas

s e

bs

s

62.{f(t)

}=

{(t 1) + (t 2) + (t 3) + } = es

s+

e2s

s+

e3s

s+ = 1

s

es

1 es63. The Laplace transform of the dierential equation is

s {y} y(0) + {y} = 5ses.

Solving for {y} we obtain{y} = 5e

s

s(s + 1)= 5es

[1s 1

s + 1

].

Thusy = 5 (t 1) 5e(t1) (t 1).


s {y} y(0) + {y} = 1s 2

ses.


{y} = 1s(s + 1)

2es

s(s + 1)=

1s 1

s + 1 2es

[1s 1

s + 1

].

Thusy = 1 et 2

[1 e(t1)

](t 1).


s {y} y(0) + 2 {y} = 1s2 es s + 1

s2.


{y} = 1s2(s + 2)

es s + 1s2(s + 2)

= 14

1s+

12

1s2

+14

1s + 2

es[14

1s+

12

1s2 1

41

s + 2

].

Thus

y = 14+

12t +

14e2t

[14+

12(t 1) 1

4e2(t1)

](t 1).


s2 {y} sy(0) y(0) + 4 {y} = 1s e

s

s.

214



{y} = 1 ss(s2 + 4)

es 1s(s2 + 4)

=14

1s 1

4s

s2 + 4 1

22

s2 + 4 es

[14

1s 1

4s

s2 + 4

].

Thus

y =14 1

4cos 2t 1

2sin 2t

[14 1

4cos 2(t 1)

](t 1).


s2 {y} sy(0) y(0) + 4 {y} = e2s 1s2 + 1

.


{y} = ss2 + 4

+ e2s[13

1s2 + 1

16

2s2 + 4

].

Thus

y = cos 2t +[13sin(t 2) 1

6sin 2(t 2)

](t 2).


s2 {y} sy(0) y(0) 5 [s {y} y(0)] + 6 {y} = es

s.


{y} = es 1s(s 2)(s 3) +

1(s 2)(s 3)

= es[16

1s 1

21

s 2 +13

1s 3

] 1

s 2 +1

s 3 .

Thus

y =[16 1

2e2(t1) +

13e3(t1)

](t 1) e2t + e3t.


s2 {y} sy(0) y(0) + {y} = es

s e

2s

s.


{y} = es[1s s

s2 + 1

] e2s

[1s s

s2 + 1

]+

1s2 + 1

.

Thusy = [1 cos(t )] (t ) [1 cos(t 2)] (t 2) + sin t.


s2 {y} sy(0) y(0) + 4[s {y} y(0)]+ 3 {y} = 1s e

2s

s e

4s

s+

e6s

s.


{y} = 13

1s 1

21

s + 1+

16

1s + 3

e2s[13

1s 1

21

s + 1+

16

1s + 3

]

e4s[13

1s 1

21

s + 1+

16

1s + 3

]+ e6s

[13

1s 1

21

s + 1+

16

1s + 3

].

215


Thus

y =13 1

2et +

16e3t

[13 1

2e(t2) +

16e3(t2)

](t 2)

[13 1

2e(t4) +

16e3(t4)

](t 4) +

[13 1

2e(t6) +

16e3(t6)

](t 6).

71. Recall from Section 3.8 that mx = kx + f(t). Now m = W/g = 32/32 = 1 slug, and 32 = 2k so thatk = 16 lb/ft. Thus, the dierential equation is x + 16x = f(t). The initial conditions are x(0) = 0, x(0) = 0.Also, since

f(t) =

{20t, 0 t < 50, t 5

and 20t = 20(t 5) + 100 we can writef(t) = 20t 20t (t 5) = 20t 20(t 5) (t 5) 100 (t 5).

The Laplace transform of the dierential equation is

s2 {x}+ 16 {x} = 20s2 20

s2e5s 100

se5s.


{x} = 20s2(s2 + 16)

20s2(s2 + 16)

e5s 100s(s2 + 16)

e5s

=(

54 1s2 5

16 4s2 + 16

)(1 e5s) (25

4 1s 25

4 ss2 + 16

)e5s.

Thus

x(t) =54t 5

16sin 4t

[54(t 5) 5

16sin 4(t 5)

](t 5)

[254 25

4cos 4(t 5)

](t 5)

=54t 5

16sin 4t 5

4t (t 5) + 5

16sin 4(t 5) (t 5) + 25

4cos 4(t 5) (t 5).

72. Recall from Section 3.8 that mx = kx + f(t). Now m = W/g = 32/32 = 1 slug, and 32 = 2k so thatk = 16 lb/ft. Thus, the dierential equation is x + 16x = f(t). The initial conditions are x(0) = 0, x(0) = 0.Also, since

f(t) ={

sin t, 0 t < 20, t 2

and sin t = sin(t 2) we can writef(t) = sin t sin(t 2) (t 2).

The Laplace transform of the dierential equation is

s2 {x}+ 16 {x} = 1s2 + 1

1s2 + 1

e2s.


{x} = 1(s2 + 16) (s2 + 1)

1(s2 + 16) (s2 + 1)

e2s

=1/15s2 + 16

+1/15s2 + 1

[ 1/15s2 + 16

+1/15s2 + 1

]e2s.

216


Thusx(t) = 1

60sin 4t +

115

sin t +160

sin 4(t 2) (t 2) 115

sin(t 2) (t 2)

=

{ 160 sin 4t + 115 sin t, 0 t < 20, t 2.


2.5dq

dt+ 12.5q = 5 (t 3).


s {q}+ 5 {q} = 2se3s.


{q} = 2s(s + 5)

e3s =(

25 1s 2

5 1s + 5

)e3s.

Thusq(t) =

25

(t 3) 25e5(t3) (t 3).


10dq

dt+ 10q = 30et 30et (t 1.5).


s {q} q0 + {q} = 3s 1

3e1.5

s 1.5 e1.5s.


{q} =(q0 32

) 1s + 1

+32 1s 1 3e

1.5

(2/5s + 1

+2/5

s 1.5)

e1.5s.

Thus

q(t) =(q0 32

)et +

32et +

65e1.5

(e(t1.5) e1.5(t1.5)

)(t 1.5).

75. (a) The dierential equation is

di

dt+ 10i = sin t + cos

(t 3

2

) (t 3

2

), i(0) = 0.


s {i}+ 10 {i} = 1s2 + 1

+se3s/2

s2 + 1.

Solving for {i} we obtain

{i} = 1(s2 + 1)(s + 10)

+s

(s2 + 1)(s + 10)e3s/2

=1

101

(1

s + 10 s

s2 + 1+

10s2 + 1

)+

1101

( 10s + 10

+10s

s2 + 1+

1s2 + 1

)e3s/2.

Thusi(t) =

1101

(e10t cos t + 10 sin t)

+1

101

[10e10(t3/2) + 10 cos

(t 3

2

)+ sin

(t 3

2

)] (t 3

2

).

217

1 2 3 4 5 6t

-0.2

0.2

i

1 2 3 4 5 6t

1

q


(b)

The maximum value of i(t) is approximately 0.1 at t = 1.7, the minimum is approximately 0.1 at 4.7.76. (a) The dierential equation is

50dq

dt+

10.01

q = E0[ (t 1) (t 3)], q(0) = 0or

50dq

dt+ 100q = E0[ (t 1) (t 3)], q(0) = 0.


50s {q}+ 100 {q} = E0(

1ses 1

se3s

).


{q} = E050

[es

s(s + 2) e

3s

s(s + 2)

]=

E050

[12

(1s 1

s + 2

)es 1

2

(1s 1

s + 2

)e3s

].

Thus

q(t) =E0100

[(1 e2(t1)

)(t 1)

(1 e2(t3)

)(t 3)

].

(b)

The maximum value of q(t) is approximately 1 at t = 3.


EId4y

dx4= w0[1 (x L/2)].

Taking the Laplace transform of both sides and using y(0) = y(0) = 0 we obtain

s4 {y} sy(0) y(0) = w0EI

1s

(1 eLs/2

).

Letting y(0) = c1 and y(0) = c2 we have

{y} = c1s3

+c2s4

+w0EI

1s5

(1 eLs/2

)so that

y(x) =12c1x

2 +16c2x

3 +124

w0EI

[x4

(x L

2

)4 (x L

2

)].

To nd c1 and c2 we compute

y(x) = c1 + c2x +12

w0EI

[x2

(x L

2

)2 (x L

2

)]

and

218


y(x) = c2 +w0EI

[x

(x L

2

) (x L

2

)].

Then y(L) = y(L) = 0 yields the system

c1 + c2L +12

w0EI

[L2

(L

2

)2]= c1 + c2L +

38

w0L2

EI= 0

c2 +w0EI

(L

2

)= c2 +

12

w0L

EI= 0.

Solving for c1 and c2 we obtain c1 = 18w0L2/EI and c2 = 12w0L/EI. Thus

y(x) =w0EI

[116

L2x2 112

Lx3 +124

x4 124

(x L

2

)4 (x L

2

)].


EId4y

dx4= w0[ (x L/3) (x 2L/3)].


s4 {y} sy(0) y(0) = w0EI

1s

(eLs/3 e2Ls/3

).


{y} = c1s3

+c2s4

+w0EI

1s5

(eLs/3 e2Ls/3

)so that

y(x) =12c1x

2 +16c2x

3 +124

w0EI

[(x L

3

)4 (x L

3

)(x 2L

3

)4 (x 2L

3

)].


y(x) = c1 + c2x +12

w0EI

[(x L

3

)2 (x L

3

)(x 2L

3

)2 (x 2L

3

)]

and

y(x) = c2 +w0EI

[(x L

3

) (x L

3

)(x 2L

3

) (x 2L

3

)].


c1 + c2L +12

w0EI

[(2L3

)2(

L

3

)2]= c1 + c2L +

16

w0L2

EI= 0

c2 +w0EI

[2L3 L

3

]= c2 +

13

w0L

EI= 0.

Solving for c1 and c2 we obtain c1 = 16w0L2/EI and c2 = 13w0L/EI. Thus

y(x) =w0EI

(112

L2x2 118

Lx3 +124

[(x L

3

)4 (x L

3

)(x 2L

3

)4 (x 2L

3

)]).


EId4y

dx4=

2w0L

[L

2 x +

(x L

2

) (x L

2

)].

219



s4 {y} sy(0) y(0) = 2w0EIL

[L

2s 1

s2+

1s2

eLs/2].


{y} = c1s3

+c2s4

+2w0EIL

[L

2s5 1

s6+

1s6

eLs/2]

so that

y(x) =12c1x

2 +16c2x

3 +2w0EIL

[L

48x4 1

120x5 +

1120

(x L

2

)5 (x L

2

)]

=12c1x

2 +16c2x

3 +w0

60EIL

[5L2

x4 x5 +(x L

2

)5 (x L

2

)].


y(x) = c1 + c2x +w0

60EIL

[30Lx2 20x3 + 20

(x L

2

)3 (x L

2

)]

and

y(x) = c2 +w0

60EIL

[60Lx 60x2 + 60

(x L

2

)2 (x L

2

)].


c1 + c2L +w0

60EIL

[30L3 20L3 + 5

2L3]= c1 + c2L +

5w0L2

24EI= 0

c2 +w0

60EIL[60L2 60L2 + 15L2] = c2 + w0L4EI = 0.

Solving for c1 and c2 we obtain c1 = w0L2/24EI and c2 = w0L/4EI. Thus

y(x) =w0L

2

48EIx2 w0L

24EIx3 +

w060EIL

[5L2

x4 x5 +(x L

2

)5 (x L

2

)].


EId4y

dx4= w0[1 (x L/2)].


s4 {y} sy(0) y(0) = w0EI

1s

(1 eLs/2

).


{y} = c1s3

+c2s4

+w0EI

1s5

(1 eLs/2

)so that

y(x) =12c1x

2 +16c2x

3 +124

w0EI

[x4

(x L

2

)4 (x L

2

)].


y(x) = c1 + c2x +12

w0EI

[x2

(x L

2

)2 (x L

2

)].

220



12c1L

2 +16c2L

3 +124

w0EI

[L4

(L

2

)4]=

12c1L

2 +16c2L

3 +5w0

128EIL4 = 0

c1 + c2L +12

w0EI

[L2

(L

2

)2]= c1 + c2L +

3w08EI

L2 = 0.

Solving for c1 and c2 we obtain c1 = 9128 w0L2/EI and c2 = 57128 w0L/EI. Thus

y(x) =w0EI

[9

256L2x2 19

256Lx3 +

124

x4 124

(x L

2

)4 (x L

2

)].

81. (a) The temperature T of the cake inside the oven is modeled by

dT

dt= k(T Tm)

where Tm is the ambient temperature of the oven. For 0 t 4, we have

Tm = 70 +300 704 0 t = 70 + 57.5t.

Hence for t 0,Tm =

{70 + 57.5t, 0 t < 4300, t 4.

In terms of the unit step function,

Tm = (70 + 57.5t)[1 (t 4)] + 300 (t 4) = 70 + 57.5t + (230 57.5t) (t 4).The initial-value problem is then

dT

dt= k[T 70 57.5t (230 57.5t) (t 4)], T (0) = 70.

(b) Let t(s) = {T (t)}. Transforming the equation, using 230 57.5t = 57.5(t 4) and Theorem 4.7, gives

st(s) 70 = k(t(s) 70

s 57.5

s2+

57.5s2

e4s)

or

t(s) =70

s k 70k

s(s k) 57.5k

s2(s k) +57.5k

s2(s k) e4s.

After using partial functions, the inverse transform is then

T (t) = 70 + 57.5(

1k

+ t 1k

ekt) 57.5

(1k

+ t 4 1k

ek(t4))

(t 4).

Of course, the obvious question is: What is k? If the cake is supposed to bake for, say, 20 minutes, thenT (20) = 300. That is,

300 = 70 + 57.5(

1k

+ 20 1k

e20k) 57.5

(1k

+ 16 1k

e16k)

.

But this equation has no physically meaningful solution. This should be no surprise since the model predictsthe asymptotic behavior T (t) 300 as t increases. Using T (20) = 299 instead, we nd, with the help of aCAS, that k 0.3.

82. In order to apply Theorem 4.7 we need the function to have the form f(t a) (t a). To accomplish thisrewrite the functions given in the forms shown below.

221


(a) 2t + 1 = 2(t 1 + 1) + 1 = 2(t 1) + 3 (b) et = et5+5 = e5et5(c) cos t = cos(t ) (d) t2 3t = (t 2)2 + (t 2) 2

83. (a) From Theorem 4.6 we have {tekti} = 1/(s ki)2. Then, using Eulers formula,

{tekti} = {t cos kt + it sin kt} = {t cos kt}+ i {t sin kt}

=1

(s ki)2 =(s + ki)2

(s2 + k2)2=

s2 k2(s2 + k2)2

+ i2ks

(s2 + k2)2.

Equating real and imaginary parts we have

{t cos kt} = s2 k2

(s2 + k2)2and {t sin kt} = 2ks

(s2 + k2)2.

(b) The Laplace transform of the dierential equation is

s2 {x}+ 2 {x} = ss2 + 2

.

Solving for {x} we obtain {x} = s/(s2 + 2)2. Thus x = (1/2)t sint.

EXERCISES 4.4Additional Operational Properties

1. {te10t} = dds

(1

s + 10

)=

1(s + 10)2

2. {t3et} = (1)3 d3

ds3

(1

s 1)

=6

(s 1)4

3. {t cos 2t} = dds

(s

s2 + 4

)=

s2 4(s2 + 4)2

4. {t sinh 3t} = dds

(3

s2 9)

=6s

(s2 9)2

5. {t2 sinh t} = d2

ds2

(1

s2 1)

=6s2 + 2(s2 1)3

6. {t2 cos t} = d2

ds2

(s

s2 + 1

)=

d

ds

(1 s2

(s2 + 1)2

)=

2s(s2 3)

(s2 + 1)3

7.{te2t sin 6t

}= d

ds

(6

(s 2)2 + 36)

=12(s 2)

[(s 2)2 + 36]2

8.{te3t cos 3t

}= d

ds

(s + 3

(s + 3)2 + 9

)=

(s + 3)2 9[(s + 3)2 + 9]2


s {y}+ {y} = 2s(s2 + 1)2

.


{y} = 2s(s + 1)(s2 + 1)2

= 12

1s + 1

12

1s2 + 1

+12

s

s2 + 1+

1(s2 + 1)2

+s

(s2 + 1)2.

222

4.4 Additional Operational Properties

Thus

y(t) = 12et 1

2sin t +

12cos t +

12(sin t t cos t) + 1

2t sin t

= 12et +

12cos t 1

2t cos t +

12t sin t.


s {y} {y} = 2(s 1)((s 1)2 + 1)2 .

Solving for {y} we obtain{y} = 2

((s 1)2 + 1)2 .

Thus

y = et sin t tet cos t.


s2 {y} sy(0) y(0) + 9 {y} = ss2 + 9

.

Letting y(0) = 2 and y(0) = 5 and solving for {y} we obtain

{y} = 2s3 + 5s2 + 19s 45

(s2 + 9)2=

2ss2 + 9

+5

s2 + 9+

s

(s2 + 9)2.

Thus

y = 2 cos 3t +53sin 3t +

16t sin 3t.


s2 {y} sy(0) y(0) + {y} = 1s2 + 1

.


{y} = s3 s2 + s(s2 + 1)2

=s

s2 + 1 1

s2 + 1+

1(s2 + 1)2

.

Thus

y = cos t sin t +(

12sin t 1

2t cos t

)= cos t 1

2sin t 1

2t cos t.


s2 {y} sy(0) y(0) + 16 {y} = {cos 4t cos 4t (t )}or by (16) of Section 4.3 in the text,

(s2 + 16) {y} = 1 + ss2 + 16

es {cos 4(t + )}

= 1 +s

s2 + 16 es {cos 4t} = 1 + s

s2 + 16 s

s2 + 16es .

Thus

{y} = 1s2 + 16

+s

(s2 + 16)2 s

(s2 + 16)2es

and

y =14sin 4t +

18t sin 4t 1

8(t ) sin 4(t ) (t ).

223

1 2 3 4 5 6t

-1

-0.5

0.5

1

y

1 2 3 4 5 6t

-4

-2

2

4

y



s2 {y} sy(0) y(0) + {y} ={1

(t

2

)+ sin t

(t

2

)}

(s2 + 1) {y} = s + 1s 1

ses/2 + es/2

{sin

(t +

2

)}or

= s +1s 1

ses/2 + es/2 {cos t}

= s +1s 1

ses/2 +

s

s2 + 1es/2.

Thus{y} = s

s2 + 1+

1s(s2 + 1)

1s(s2 + 1)

es/2 +s

(s2 + 1)2es/2

=s

s2 + 1+

1s s

s2 + 1(

1s s

s2 + 1

)es/2 +

s

(s2 + 1)2es/2

=1s(

1s s

s2 + 1

)es/2 +

s

(s2 + 1)2es/2

andy = 1

[1 cos

(t

2

)] (t

2

)+

12

(t

2

)sin

(t

2

) (t

2

)= 1 (1 sin t)

(t

2

) 1

2

(t

2

)cos t

(t

2

).

15. 16.

17. From (7) of Section 4.2 in the text along with Theorem 4.8,

{ty} = dds

{y} = dds

[s2Y (s) sy(0) y(0)] = s2 dYds

2sY + y(0),so that the transform of the given second-order dierential equation is the linear rst-order dierential equationin Y (s):

s2Y + 3sY = 4s3

or Y +3sY = 4

s5.

The solution of the latter equation is Y (s) = 4/s4 + c/s3, so

y(t) = {Y (s)} = 23t3 +

c

2t2.

18. From Theorem 4.8 in the text

{ty} = dds

{y} = dds

[sY (s) y(0)] = s dYds

Yso that the transform of the given second-order dierential equation is the linear rst-order dierential equationin Y (s):

Y +(

3s 2s

)Y = 10

s.

224


Using the integrating factor s3es2, the last equation yields

Y (s) =5s3

+c

s3es

2.

But if Y (s) is the Laplace transform of a piecewise-continuous function of exponential order, we must have, inview of Theorem 4.5, lims Y (s) = 0. In order to obtain this condition we require c = 0. Hence

y(t) ={

5s3

}=

52t2.

19.{1 t3} = 1

s

3!s4

=6s5

20.{t2 tet} = 2

s3(s 1)2

21.{et et cos t} = s 1

(s + 1) [(s 1)2 + 1] 22.{e2t sin t} = 1

(s 2)(s2 + 1)

23.{ t

0

e d

}=

1s

{et} = 1s(s 1)

24.{ t

0

cos d}

=1s

{cos t} = ss(s2 + 1)

=1

s2 + 1

25.{ t

0

e cos d}

=1s

{et cos t

}=

1s

s + 1(s + 1)2 + 1

=s + 1

s (s2 + 2s + 2)

26.{ t

0

sin d}

=1s

{t sin t} = 1s

( d

ds

1s2 + 1

)= 1

s

2s(s2 + 1)2

=2

(s2 + 1)2

27.{ t

0

et d}

= {t} {et} = 1s2(s 1)

28.{ t

0

sin cos(t ) d}

= {sin t} {cos t} = s(s2 + 1)2

29.{t

t0

sin d}

= dds

{ t0

sin d}

= dds

(1s

1s2 + 1

)=

3s2 + 1s2 (s2 + 1)2

30.{t

t0

ed}

= dds

{ t0

ed}

= dds

(1s

1(s + 1)2

)=

3s + 1s2(s + 1)3

31.{

1s(s 1)

}=

{1/(s 1)

s

}= t0

ed = et 1

32.{

1s2(s 1)

}=

{1/s(s 1)

s

}= t0

(e 1)d = et t 1

33.{

1s3(s 1)

}=

{1/s2(s 1)

s

}= t0

(e 1)d = et 12t2 t 1

34. Using{

1(s a)2

}= teat, (8) in the text gives

{1

s(s a)2}

= t0

ea d =1a2

(ateat eat + 1).

35. (a) The result in (4) in the text is {F (s)G(s)} = f g, so identify

F (s) =2k3

(s2 + k2)2and G(s) =

4ss2 + k2

.

225

ty

5 10 1550

50


Thenf(t) = sin kt kt cos kt and g(t) = 4 cos kt

so {8k3s

(s2 + k2)3

}= {F (s)G(s)} = f g = 4

t0

f()g(t )dt

= 4 t0

(sin k k cos k) cos k(t )d.

Using a CAS to evaluate the integral we get{8k3s

(s2 + k2)3

}= t sin kt kt2 cos kt.

(b) Observe from part (a) that

{t(sin kt kt cos kt)} = 8k3s

(s2 + k2)3,

and from Theorem 4.8 that{tf(t)

}= F (s). We saw in (5) in the text that

{sin kt kt cos kt} = 2k3/(s2 + k2)2,so {

t(sin kt kt cos kt)} = dds

2k3

(s2 + k2)2=

8k3s(s2 + k2)3

.


s2 {y}+ {y} = 1(s2 + 1)

+2s

(s2 + 1)2.

Thus

{y} = 1(s2 + 1)2

+2s

(s2 + 1)3

and, using Problem 35 with k = 1,

y =12(sin t t cos t) + 1

4(t sin t t2 cos t).

37. The Laplace transform of the given equation is

{f}+ {t} {f} = {t}.

Solving for {f} we obtain {f} = 1s2 + 1

. Thus, f(t) = sin t.


{f} = {2t} 4 {sin t} {f}.Solving for {f} we obtain

{f} = 2s2 + 2

s2(s2 + 5)=

25

1s2

+8

55

5

s2 + 5.

Thusf(t) =

25t +

855sin

5 t.


{f} = {tet}+ {t} {f}.226

5.Laplace Casos Especiais.1234.Sol

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Transcript of 5.Laplace Casos Especiais.1234.Sol