Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions

8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions

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EXERCISES 16.3 11

y(t) =L1

As+ B

s2 + 9 +

1

(s i)2(s2 + 9)

=L1

As+ B

s2 + 9 i/32

s

i

+ 1/8

(s

i)2

+is/32 5/32

s2 + 9 =L1

Cs + Ds2 + 9

i/32s i+

1/8(s i)2

= Ccos 3t +

D3

sin3t i32

eti + t8

eti.

If we take imaginary parts, we get y(t) =Ccos 3t + Esin 3t 132

cos t + t

8sin t.

36. Assuming that the solution satisfies the conditions of Corollary 16.5.1, we take Laplace transforms ofboth sides of the differential equation and use the initial conditions,

[s3Y s2(1) + 2] 3[s2Y s(1)] + 3[sY 1] Y = 2(s 1)3 .

We solve this for the transform Y(s),

Y(s) =

s2

3s + 1

s3 3s2 + 3s 1+ 2

(s 1)3(s3 3s2 + 3s 1)=s2

3s + 1

(s 1)3 + 2

(s 1)6 .The inverse transform of this function is the solution of the initial-value problem

y(t) =L1

s2 3s + 1(s 1)3 +

2

(s 1)6

= L1

1

s 1 1

(s 1)2 1

(s 1)3 + 2

(s 1)6

=etL1

1

s 1

s2 1

s3+

2

s6

= et

1 t t

2

2 +

t5

60

.

38. The initial-value problem is

1

5

d2x

dt2 + 10x= 0 = x+ 50x= 0, x(0) = 0.03, x(0) = 0.

Assuming that the solution satisfies the conditions of Corollary 16.5.1, we take Laplace transforms ofboth sides of the differential equation and use the initial conditions,

[s2X+ 0.03s] + 50X= 0.

We solve this for the transform X(s),

X(s) = 0.03ss2 + 50

.

The inverse transform of this function is the solution of the initial-value problem

x(t) = L1 0.03s

s2 + 50

=0.03cos5

2t m.

40. The initial-value problem is

1

5

d2x

dt2 +

3

2

dx

dt + 10x= 4sin10t = 2x+ 15x+ 100x= 40sin10t, x(0) = 0, x(0) = 0.

Assuming that the solution satisfies the conditions of Corollary 16.5.1, we take Laplace transforms ofboth sides of the differential equation and use the initial conditions,

2[s2X] + 15[sX] + 100X= 400

s2 + 100.



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14 EXERCISES 16.4


y(t) =L1

s + 6

(s+ 1)(s + 3)+

1 + es

(s2 + 1)(s + 1)(s+ 3)

=L1

5/2

s+ 1 3/2

s + 3+

1/4

s + 1 1/20

s+ 3+s/5 + 1/10

s2 + 1

(1 + es)

=52

et 32

e3t + 120

5et e3t 4cos t+ 2 sin t

+ 1

20

5e(t) e3(t) 4cos(t ) + 2 sin (t )

h(t )

=11

4et 31

20e3t 1

5cos t +

1

10sin t +

1

20

5et e3(t) + 4 cos t 2sin t

h(t ).

8. When we take Laplace transforms of both sides of the differential equation and use the initial conditions,

[s2Y] + 2[sY] + 5Y = L{4[h(t) h(t 1)] 4[h(t 1) h(t 2)]}= L{4 8h(t 1) + 4h(t 2)}=

4

s 8e

s

s +

4e2s

s .


Y(s) = 4

s(s2 + 2s + 5)(1 2es + e2s).


y(t) = L1

4

s(s2 + 2s + 5)(1 2es + e2s)

= 4L1

1/5

s s/5 + 2/5

s2 + 2s + 5

(1 2es + e2s)

=4

5L1

1

s (s+ 1) + 1

(s + 1)2 + 4

(1 2es + e2s)

=4

5

1 et

cos2t +

1

2sin 2t

8

5

1 e(t1)

cos2(t 1) +1

2sin 2(t 1)

h(t 1)

+

4

5

1 e(t

2)

cos2(t 2) +1

2sin2(t 2)h(t 2)=

2

5

2 et (2cos2t + sin 2t) +4

5

2 + e1t [2cos2(t 1) + sin 2(t 1)]h(t 1)+

2

5

2 e2t [2cos2(t 2) + sin 2(t 2)]h(t 2).

10. When we take Laplace transforms of both sides of the differential equation and use the initial conditions,

[s2Y s(2)] + 16Y = 11 e2sL{t[h(t) h(t 1)] + (2 t)[h(t 1) h(t 2)]}

= 1

1 e2sL{t + (2 2t)h(t 1) + (t 2)h(t 2)}

= 1

1 e2s 1

s2+ es

L{2

2(t+ 1)}

+ e2s

L{t}

= 1

1 e2s

1

s2 2e

s

s2 +

e2s

s2

=

(1 es)2s2(1 es)(1 + es) =

1 ess2(1 + es)

.


Y(s) = 2s

s2 + 16+

1 ess2(s2 + 16)(1 + es)

.



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EXERCISES 16.4 15

y(t) =L1

2s

s2 + 16+

1 ess2(s2 + 16)(1 + es)

=L1 2s

s2 + 16+

1/16

s2 1/16

s2 + 16 (1 es)

n=0

(1)nens=L1

2s

s2 + 16+

1

16

1

s2 1

s2 + 16

n=0

(1)nens +n=0

(1)n+1e(n+1)s

= 2cos 4t + 1

16

n=0

(1)n

(t n) 14

sin 4(t n)

h(t n)

+ 1

16

n=0

(1)n+1

(t n 1) 14

sin 4(t n 1)

h(t n 1)

= 2cos 4t + 1

64

n=0

(1)n [4(t n) sin4(t n)] h(t n)

+

1

64

n=0

(1)n+1

[4(t n 1) sin4(t n 1)] h(t n 1).

12. The initial-value problem for displacement of the mass from its equilibrium position is

1

10

d2x

dt2 + 40x= 100h(t 4), x(0) = 1

10, x(0) =2.

We write the differential equation in the form

d2x

dt2+ 400x= 1000h(t 4),

and take Laplace transforms,

s2

X s

10+ 2

+ 400X=

1000e4s

s .


X(s) =s/10 2s2 + 400

+ 1000e4s

s(s2 + 400).


x(t) =L1

s/10 2s2 + 400

+ 1000e4s

s(s2 + 400)

=L1

s/10 2s2 + 400

+5

2

1

s s

s2 + 400

e4s

=

1

10cos20t 1

10sin20t +

5

2 [1 cos20(t 4)]h(t 4) m.14. The initial-value problem for displacement of the mass from its equilibrium position is

1

10

d2x

dt2 + 5

dx

dt + 40x= 100h(t 4), x(0) = 1

10, x(0) =2.

We write the differential equation in the form

d2x

dt2 + 50

dx

dt+ 400x= 1000h(t 4),


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EXERCISES 16.4 17


2d2x

dt2 + 80

dx

dt+ 512x= (t), x(0) = 0, x(0) = 0.

When we take Laplace transforms,

2[s2X] + 80[sX] + 512X= 1.


X(s) = 1

2s2 + 80s + 512=

1

2(s2 + 40s + 256).


x(t) =L1

1

2(s2 + 40s + 256)

=

1

2L1

1

(s + 20)2 144

=e20t

2 L1

1

s2 144

=e20t

2 L1

1/24s + 12

+ 1/24

s 12

=

e20t

48e12t + e12t=

1

48

e8t

e32t

m.


2d2x

dt2+ 512x=(t t0), x(0) =x0, x(0) = 0.


2[s2X x0s] + 512X= et0s.We solve this for the transform X(s),

X(s) = 2x0s

2s2 + 512+

et0s

2s2 + 512=

x0s

s2 + 256+

et0s

2(s2 + 256).


x(t) =L1

x0s

s2 + 256+

et0s

2(s2 + 256)

= x0cos 16t +

1

32sin 16(t t0) h(t t0) m.


2d2x

dt2+ 512x=(t t0), x(0) =x0, x(0) =v0.


2[s2Xx0s v0] + 512X= et0s.


X(s) =2x0s + 2v0

2s2 + 512 +

et0s

2s2 + 512=

x0s+ v0s2 + 256

+ et0s

2(s2 + 256).


x(t) = L1

x0s+ v0s2 + 256

+ et0s

2(s2 + 256)

= x0cos 16t +

v016

sin16t + 1

32sin16(t t0) h(t t0) m.



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18 EXERCISES 16.4

d2x

dt2+ 100x=

n=0

(t n), x(0) = 0, x(0) = 0.


[s2X] + 100X=

n=0

ens.


X(s) =

n=0

ens

s2 + 100.


x(t) =L1n=0

ens

s2 + 100

=

1

10

n=0

sin10(t n) h(t n) m.


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EXERCISES 16.5 21

Heaviside function have continuous derivatives of all orders. Consider then, the Heaviside term, less theleading constant, f(x) = (x L/3)3h(x L/3). Clearly,

limxL/3+

f(x) = limxL/3

f(x).

Sincef(x) = 3(x L/3)2h(x L/3) andf(x) = 6(x L/3)h(x L/3), we also see thatlim

xL/3+ f(x) = lim

xL/3 f(x) and lim

xL/3+ f(x) = lim

xL/3 f(x).

On the other hand, since f(x) = 6h(x L/3),lim

xL/3+f(x) = 6 and lim

xL/3f(x) = 0.

8. The boundary-value problem for deflections of the beam is

d4y

dx4 =P

EI(x L/2), y(0) =y(0) = 0, y(L) = y(L) = 0.

If we set y (0) =A and y (0) =B, and take Laplace transforms, we obtain

s4Y

As

B=

P

EIeLs/2.


Y(s) = A

s3+

B

s4 P e

Ls/2

EI s4 .


y(x) =Ax2

2 +

B x3

6 P

6EI(x L/2)3h(x L/2).

The boundary conditions atx = L require

0 = y(L) =A+ BL PEI

L

2 , 0 = y(L) =B P

EI.

The solution of these equations is A = P L2EI

andB = P

EI.

Thus,

y(x) =P Lx2

4EI +

P x3

6EI P

6EI(x L/2)3h(x L/2).


d4y

dx4 =P

EI(x L/2), y(0) =y(0) = 0, y(L) =y(L) = 0.


s4

Y As2

B = P

EIeLs/2

.


Y(s) = A

s2+

B

s4 P e

Ls/2

EI s4 .


y(x) =Ax+B x3

6 P

6EI(x L/2)3h(x L/2).


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22 EXERCISES 16.5

The boundary conditions atx = L require

0 = y(L) = AL +B L3

6 P

6EI

L

2

3, 0 =y(L) =BL P

EI

L

2

.

The solution of these equations is A = P L2

16EI and B =

P

2EI.

Thus,

y(x) =P L2x

16EI +

P x3

12EI P

6EI(x L/2)3h(x L/2).

12. In this situation, the beam will rotate until it is vertical, hanging from the pin atx = 0. The boundary-value problem for deflections of the beam is

d4y

dx4 = mg

EI L[h(x) h(x L)], y(0) =y (0) = 0, y(L) =y(L) = 0.


s4Y As2 B=mgEI L

1

s e

Ls

s = mg

EI L 1 eLs

s .We solve this for the transform Y(s),

Y(s) = A

s2+

B

s4 mg

EI L

1 eLs

s5

.


y(x) =Ax+B x3

6 mgx

4

24EI L+

mg (x L)424EI L

h(x L).

Since the last term contributes nothing to the solution, we drop it from further consideration. Theboundary conditions at x = L require

0 = y(L) =BL

mgL

2EI

, 0 = y(L) =B

mg

EI

.

These equations give conflicting values for B. Hence, the boundary-value problem does not give thephysical solution.


d4y

dx4 = mg

EI L[h(x) h(x L)], y(0) =y(0) = 0, y(L) =y(L) = 0.


s4Y As2 B=mgEI L

1

s e

Ls

s

= mg

EI L

1 eLs

s

.


Y(s) = A

s2+

B

s4 mg

EI L

1 eLs

s5

.


y(x) =Ax+B x3

6 mgx

4

24EI L+

mg (x L)424EI L

h(x L).

Since the last term contributes nothing to the solution, we drop it from further consideration. Theboundary conditions at x = L require


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EXERCISES 16.5 23

0 = y(L) =AL+B L3

6 mgL

3

24EI, 0 = y(L) =BLmg L

2EI.

The solution of these equations is A = mgL2

24EI andB =

mg

2EI. Thus,

y(x) =

mgL2x

24EI

+mgx3

12EI mgx4

24EI L

=

mg

24EI L

(x4

2Lx3 +L3x).

Maximum deflection occurs at x = L/2,

y(L/2) = mg24EI L

L

2

4 2L

L

2

3+ L3

L

2

=5mgL

3

384EI.

For this to be less than L/100,

5mgL3

384EI

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Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions