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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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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.
We solve this for the transform X(s),
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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14 EXERCISES 16.4
The inverse transform of this function is the solution of the initial-value problem
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 .
We solve this for the transform Y(s),
Y(s) = 4
s(s2 + 2s + 5)(1 2es + e2s).
The inverse transform of this function is the solution of the initial-value problem
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)
.
We solve this for the transform Y(s),
Y(s) = 2s
s2 + 16+
1 ess2(s2 + 16)(1 + es)
.
The inverse transform of this function is the solution of the initial-value problem
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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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 .
We solve this for the transform X(s),
X(s) =s/10 2s2 + 400
+ 1000e4s
s(s2 + 400).
The inverse transform of this function is the solution of the initial-value problem
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),
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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EXERCISES 16.4 17
18. The initial-value problem for displacement of the mass from its equilibrium position is
2d2x
dt2 + 80
dx
dt+ 512x= (t), x(0) = 0, x(0) = 0.
When we take Laplace transforms,
2[s2X] + 80[sX] + 512X= 1.
We solve this for the transform X(s),
X(s) = 1
2s2 + 80s + 512=
1
2(s2 + 40s + 256).
The inverse transform of this function is the solution of the initial-value problem
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.
20. The initial-value problem for displacement of the mass from its equilibrium position is
2d2x
dt2+ 512x=(t t0), x(0) =x0, x(0) = 0.
When we take Laplace transforms,
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).
The inverse transform of this function is the solution of the initial-value problem
x(t) =L1
x0s
s2 + 256+
et0s
2(s2 + 256)
= x0cos 16t +
1
32sin 16(t t0) h(t t0) m.
22. The initial-value problem for displacement of the mass from its equilibrium position is
2d2x
dt2+ 512x=(t t0), x(0) =x0, x(0) =v0.
When we take Laplace transforms,
2[s2Xx0s v0] + 512X= et0s.
We solve this for the transform X(s),
X(s) =2x0s + 2v0
2s2 + 512 +
et0s
2s2 + 512=
x0s+ v0s2 + 256
+ et0s
2(s2 + 256).
The inverse transform of this function is the solution of the initial-value problem
x(t) = L1
x0s+ v0s2 + 256
+ et0s
2(s2 + 256)
= x0cos 16t +
v016
sin16t + 1
32sin16(t t0) h(t t0) m.
24. The initial-value problem for displacement of the mass from its equilibrium position is
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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18 EXERCISES 16.4
d2x
dt2+ 100x=
n=0
(t n), x(0) = 0, x(0) = 0.
When we take Laplace transforms,
[s2X] + 100X=
n=0
ens.
We solve this for the transform X(s),
X(s) =
n=0
ens
s2 + 100.
The inverse transform of this function is the solution of the initial-value problem
x(t) =L1n=0
ens
s2 + 100
=
1
10
n=0
sin10(t n) h(t n) m.
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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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.
We solve this for the transform Y(s),
Y(s) = A
s3+
B
s4 P e
Ls/2
EI s4 .
The inverse transform of this function is the solution of the initial-value problem
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).
10. 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
s4
Y As2
B = P
EIeLs/2
.
We solve this for the transform Y(s),
Y(s) = A
s2+
B
s4 P e
Ls/2
EI s4 .
The inverse transform of this function is the solution of the initial-value problem
y(x) =Ax+B x3
6 P
6EI(x L/2)3h(x L/2).
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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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.
If we set y (0) =A and y (0) =B, and take Laplace transforms, we obtain
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
.
The inverse transform of this function is the solution of the initial-value problem
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.
14. 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.
If we set y (0) =A and y (0) =B, and take Laplace transforms, we obtain
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
.
The inverse transform of this function is the solution of the initial-value problem
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
8/10/2019 Calculus for Engineers: Chapter 16 - Laplace Transforms - Solutions
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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