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Transcript of Engineering Mathematics II (2M03) Tutorial 8dmpeli.math.mcmaster.ca/.../HomeAss/tutorial8.pdf ·...
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Engineering Mathematics II (2M03)Tutorial 8
Marina ChugunovaDepartment of Math. & Stat., office: HH403
e-mail: [email protected]
office hours: Math Help Centre, Thursday 1:30 - 3:30
November 1-2, 2007
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 40)
Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.
Solution:
2
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 40)
Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
3
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 40)
Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
Translation theorem:
L{g(t)U(t− a)} = e−asL{g(t + a)}
4
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 40)
Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
Translation theorem:
L{g(t)U(t− a)} = e−asL{g(t + a)}
a = 1, g(t) = 3t + 1, g(t + 1) = 3(t + 1) + 1
5
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 40)
Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
Translation theorem:
L{g(t)U(t− a)} = e−asL{g(t + a)}
a = 1, g(t) = 3t + 1, g(t + 1) = 3(t + 1) + 1
F (s) = L{(3t + 1)U(t− 1)} = e−sL{(3(t + 1) + 1)} = e−sL{(3t + 4)}
6
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 40)
Find the Laplace transform F (s) = L{(3t + 1)U(t− 1)}.
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
Translation theorem:
L{g(t)U(t− a)} = e−asL{g(t + a)}
a = 1, g(t) = 3t + 1, g(t + 1) = 3(t + 1) + 1
F (s) = L{(3t + 1)U(t− 1)} = e−sL{(3(t + 1) + 1)} = e−sL{(3t + 4)}
F (s) = 3e−s 1
s2+ 4e−s1
s
7
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 60)
Find the Laplace transform of the function:
f (t) =
{sin t, 0 ≤ t < 2π0, t ≥ 2π
Solution:
8
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 60)
Find the Laplace transform of the function:
f (t) =
{sin t, 0 ≤ t < 2π0, t ≥ 2π
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
1− U(t− a) =
{1, 0 ≤ t < a0, t ≥ a
9
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 60)
Find the Laplace transform of the function:
f (t) =
{sin t, 0 ≤ t < 2π0, t ≥ 2π
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
1− U(t− a) =
{1, 0 ≤ t < a0, t ≥ a
f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)
10
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 60)
Find the Laplace transform of the function:
f (t) =
{sin t, 0 ≤ t < 2π0, t ≥ 2π
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
1− U(t− a) =
{1, 0 ≤ t < a0, t ≥ a
f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)
F (s) = L{f (t)} =1
s2 + 1− e−2πsL{sin(t + 2π)}
11
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 60)
Find the Laplace transform of the function:
f (t) =
{sin t, 0 ≤ t < 2π0, t ≥ 2π
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
1− U(t− a) =
{1, 0 ≤ t < a0, t ≥ a
f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)
F (s) = L{f (t)} =1
s2 + 1− e−2πsL{sin(t + 2π)}
F (s) = L{f (t)} =1
s2 + 1− e−2πsL{sin t} =
1
s2 + 1− e−2πs 1
s2 + 1
12
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 60)
Find the Laplace transform of the function:
f (t) =
{sin t, 0 ≤ t < 2π0, t ≥ 2π
Solution:
U(t− a) =
{0, 0 ≤ t < a1, t ≥ a
1− U(t− a) =
{1, 0 ≤ t < a0, t ≥ a
f (t) = sin t(1− U(t− 2π)) = sin t− sin tU(t− 2π)
F (s) = L{f (t)} =1
s2 + 1− e−2πsL{sin(t + 2π)}
F (s) = L{f (t)} =1
s2 + 1− e−2πsL{sin t} =
1
s2 + 1− e−2πs 1
s2 + 1
F (s) =1
s2 + 1(1− e−2πs)
13
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =
{1, 0 ≤ t < 10, t ≥ 1
Solution:
14
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =
{1, 0 ≤ t < 10, t ≥ 1
Solution:
f (t) = 1− U(t− 1), L{f (t)} =1
s− L{U(t− 1)} =
1
s− e−s
s
15
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =
{1, 0 ≤ t < 10, t ≥ 1
Solution:
f (t) = 1− U(t− 1), L{f (t)} =1
s− L{U(t− 1)} =
1
s− e−s
s
s2Y (s) + 1 + 4Y (s) =1
s(1− e−s), Y (s)[s2 + 4] =
1
s(1− e−s)− 1
16
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =
{1, 0 ≤ t < 10, t ≥ 1
Solution:
f (t) = 1− U(t− 1), L{f (t)} =1
s− L{U(t− 1)} =
1
s− e−s
s
s2Y (s) + 1 + 4Y (s) =1
s(1− e−s), Y (s)[s2 + 4] =
1
s(1− e−s)− 1
Y (s) =1
s(s2 + 4)− e−s
s(s2 + 4)− 1
s2 + 4
17
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =
{1, 0 ≤ t < 10, t ≥ 1
Solution:
f (t) = 1− U(t− 1), L{f (t)} =1
s− L{U(t− 1)} =
1
s− e−s
s
s2Y (s) + 1 + 4Y (s) =1
s(1− e−s), Y (s)[s2 + 4] =
1
s(1− e−s)− 1
Y (s) =1
s(s2 + 4)− e−s
s(s2 + 4)− 1
s2 + 4
Y (s) = 1/41
s− 1/4
s
s2 + 4− 1/4
e−s
s+ 1/4
e−ss
s2 + 4− 1/2
2
s2 + 4
18
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The Laplace Transform (4.3 Translation Theorems)
Problem (4.3: 66)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + 4y = f (t), y(0) = 0, y′(0) = −1, f(t) =
{1, 0 ≤ t < 10, t ≥ 1
Solution:
f (t) = 1− U(t− 1), L{f (t)} =1
s− L{U(t− 1)} =
1
s− e−s
s
s2Y (s) + 1 + 4Y (s) =1
s(1− e−s), Y (s)[s2 + 4] =
1
s(1− e−s)− 1
Y (s) =1
s(s2 + 4)− e−s
s(s2 + 4)− 1
s2 + 4
Y (s) = 1/41
s− 1/4
s
s2 + 4− 1/4
e−s
s+ 1/4
e−ss
s2 + 4− 1/2
2
s2 + 4
y(t) = 1/4− 1/4 cos(2t)− 1/2 sin(2t)− 1/4U(t− 1) + 1/4 cos(2(t− 1))U(t− 1)
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 6)Find the Laplace transform F (s) of f (t) = t2 cos t.
Solution:
20
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 6)Find the Laplace transform F (s) of f (t) = t2 cos t.
Solution:
Derivatives of the Laplace transform
L{tny(t)} = (−1)ndn
dsnY (s)
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 6)Find the Laplace transform F (s) of f (t) = t2 cos t.
Solution:
Derivatives of the Laplace transform
L{tny(t)} = (−1)ndn
dsnY (s)
F (s) = L{t2 cos t} = (−1)2d2
ds2L{cos t} =
d2
ds2
(s
s2 + 1
)=
2s3 − 6s
(s2 + 1)3
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
23
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
24
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
25
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
Y (s) = −e−π2s 1
s(s2 + 1)2+
1
s
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 14)Use the Laplace transform Y (s) to solve the initial value problem
y′′ + y = f (t), y(0) = 1, y′(0) = 0, f(t) =
{1, 0 ≤ t < π/2sin t, t ≥ π/2
Solution:
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
Y (s) = −e−π2s 1
s(s2 + 1)2+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− s
(s2 + 1)2
)+
1
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The Laplace Transform (4.4 Additional Operational Properties )
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
Y (s) = −e−π2s 1
s(s2 + 1)2+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− s
(s2 + 1)2
)+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− 1/2(−1)
d
ds
(1
s2 + 1
))+
1
s
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The Laplace Transform (4.4 Additional Operational Properties )
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
Y (s) = −e−π2s 1
s(s2 + 1)2+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− s
(s2 + 1)2
)+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− 1/2(−1)
d
ds
(1
s2 + 1
))+
1
s
y(t) = U(t− π/2)[−1 + cos(t− π/2) + 1/2(t− π/2) sin(t− π/2)] + 1
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The Laplace Transform (4.4 Additional Operational Properties )
f (t) = U(t− π/2) sin t + (1− U(t− π/2)) · 1 = U(t− π/2) sin t + 1− U(t− π/2)
L{g(t)U(t− a)} = e−asL{g(t + a)}
F (s) = L{f (t)} = e−π2sL{sin(t + π/2)} +
1
s− 1
se−
π2s
F (s) = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
s2Y (s)− s + Y (s) = Y (s)[s2 + 1]− s = e−π2s s
s2 + 1+
1
s− 1
se−
π2s
Y (s) = −e−π2s 1
s(s2 + 1)2+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− s
(s2 + 1)2
)+
1
s
Y (s) = −e−π2s
(1
s− s
s2 + 1− 1/2(−1)
d
ds
(1
s2 + 1
))+
1
s
y(t) = U(t− π/2)[−1 + cos(t− π/2) + 1/2(t− π/2) sin(t− π/2)] + 1
y(t) = U(t− π/2)[−1 + sin t− 1/2(t− π/2) cos t] + 1
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 26)Find
L{∫ t
0
sin τ τdτ
}Solution:
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 26)Find
L{∫ t
0
sin τ τdτ
}Solution:
Transform of the integral
L{∫ t
0
f (τ ) dτ
}=
F (s)
s
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 26)Find
L{∫ t
0
sin τ τdτ
}Solution:
Transform of the integral
L{∫ t
0
f (τ ) dτ
}=
F (s)
s
L{∫ t
0
sin τ τdτ
}=
1
sL{t sin t}
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 26)Find
L{∫ t
0
sin τ τdτ
}Solution:
Transform of the integral
L{∫ t
0
f (τ ) dτ
}=
F (s)
s
L{∫ t
0
sin τ τdτ
}=
1
sL{t sin t} = −1
s
d
ds
(1
s2 + 1
)
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 26)Find
L{∫ t
0
sin τ τdτ
}Solution:
Transform of the integral
L{∫ t
0
f (τ ) dτ
}=
F (s)
s
L{∫ t
0
sin τ τdτ
}=
1
sL{t sin t} = −1
s
d
ds
(1
s2 + 1
)=
2
(s2 + 1)2
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 38)Use the Laplace transform to solve the integral equation
f (t) = 2t− 4
∫ t
0
sin τ f(t− τ ) dτ
Solution:
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 38)Use the Laplace transform to solve the integral equation
f (t) = 2t− 4
∫ t
0
sin τ f(t− τ ) dτ
Solution:By the Convolution theorem
F (s) = 21
s2− 4
(1
s2 + 1F (s)
)
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 38)Use the Laplace transform to solve the integral equation
f (t) = 2t− 4
∫ t
0
sin τ f(t− τ ) dτ
Solution:By the Convolution theorem
F (s) = 21
s2− 4
(1
s2 + 1F (s)
)F (s) =
2(s2 + 1)
s2(s2 + 5)=
2
s2 + 5+
2/5
s2− 2/5
s2 + 5=
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 38)Use the Laplace transform to solve the integral equation
f (t) = 2t− 4
∫ t
0
sin τ f(t− τ ) dτ
Solution:By the Convolution theorem
F (s) = 21
s2− 4
(1
s2 + 1F (s)
)F (s) =
2(s2 + 1)
s2(s2 + 5)=
2
s2 + 5+
2/5
s2− 2/5
s2 + 5=
=8
5√
5
√5
s2 + 5+
2
5
1
s2
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The Laplace Transform (4.4 Additional Operational Properties )
Problem (4.4: 38)Use the Laplace transform to solve the integral equation
f (t) = 2t− 4
∫ t
0
sin τ f(t− τ ) dτ
Solution:By the Convolution theorem
F (s) = 21
s2− 4
(1
s2 + 1F (s)
)F (s) =
2(s2 + 1)
s2(s2 + 5)=
2
s2 + 5+
2/5
s2− 2/5
s2 + 5=
=8
5√
5
√5
s2 + 5+
2
5
1
s2
f (t) =8
5√
5sin 5t +
2
5t
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
sX(s)− 1 = 2Y (s) +1
s− 1, sY (s)− 1 = 8X(s)− 1
s2
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
sX(s)− 1 = 2Y (s) +1
s− 1, sY (s)− 1 = 8X(s)− 1
s2
Y (s) = 24
s2 − 16+
s
s2 − 16+
8
(s− 1)(s2 − 16)− 1
s(s2 − 16)
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
sX(s)− 1 = 2Y (s) +1
s− 1, sY (s)− 1 = 8X(s)− 1
s2
Y (s) = 24
s2 − 16+
s
s2 − 16+
8
(s− 1)(s2 − 16)− 1
s(s2 − 16)
Y (s) = 24
s2 − 16+
s
s2 − 16− 8/15
1
s− 1+ 1/3
1
s− 4+ 1/5
1
s + 4− 1/16
(s
s2 − 16− 1
s
)
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
sX(s)− 1 = 2Y (s) +1
s− 1, sY (s)− 1 = 8X(s)− 1
s2
Y (s) = 24
s2 − 16+
s
s2 − 16+
8
(s− 1)(s2 − 16)− 1
s(s2 − 16)
Y (s) = 24
s2 − 16+
s
s2 − 16− 8/15
1
s− 1+ 1/3
1
s− 4+ 1/5
1
s + 4− 1/16
(s
s2 − 16− 1
s
)y(t) = 2 sinh 4t +
15
16cosh 4t +
1
16− 8
15et +
1
3e4t +
1
5e−4t
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
sX(s)− 1 = 2Y (s) +1
s− 1, sY (s)− 1 = 8X(s)− 1
s2
Y (s) = 24
s2 − 16+
s
s2 − 16+
8
(s− 1)(s2 − 16)− 1
s(s2 − 16)
Y (s) = 24
s2 − 16+
s
s2 − 16− 8/15
1
s− 1+ 1/3
1
s− 4+ 1/5
1
s + 4− 1/16
(s
s2 − 16− 1
s
)y(t) = 2 sinh 4t +
15
16cosh 4t +
1
16− 8
15et +
1
3e4t +
1
5e−4t
x(t) = 1/8(y′(t) + t), y′(t) = 8 cosh 4t +15
4sinh 4t− 8
15et +
4
3e4t − 4
5e−4t
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The Laplace Transform (4.6 Systems of Linear Differential Equations )
Problem (4.6: 2)Use the Laplace transform to solve the system:
dx
dt= 2y + et,
dy
dt= 8x− t, x(0) = 1, y(0) = 1
Solution:
sX(s)− 1 = 2Y (s) +1
s− 1, sY (s)− 1 = 8X(s)− 1
s2
Y (s) = 24
s2 − 16+
s
s2 − 16+
8
(s− 1)(s2 − 16)− 1
s(s2 − 16)
Y (s) = 24
s2 − 16+
s
s2 − 16− 8/15
1
s− 1+ 1/3
1
s− 4+ 1/5
1
s + 4− 1/16
(s
s2 − 16− 1
s
)y(t) = 2 sinh 4t +
15
16cosh 4t +
1
16− 8
15et +
1
3e4t +
1
5e−4t
x(t) = 1/8(y′(t) + t), y′(t) = 8 cosh 4t +15
4sinh 4t− 8
15et +
4
3e4t − 4
5e−4t
x(t) = cosh 4t +15
32sinh 4t− 1
15et +
1
6e4t − 1
10e−4t +
1
8t
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See you next week :-) !
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