8. The specific solution of a state space equation is given below. Express the solution Q terms of sines and cosines using the Euler representation of the complex exponential. [i] = 0.3 je'¹'|'|| -0.3 je¯¹¹[-';}] You calculated that Q = qiscos(1) + 916sin(1)
8. The specific solution of a state space equation is given below. Express the solution Q terms of sines and cosines using the Euler representation of the complex exponential. [i] = 0.3 je'¹'|'|| -0.3 je¯¹¹[-';}] You calculated that Q = qiscos(1) + 916sin(1)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
I need questions 8 and 9
![.Telkom-Monate LTE
11:04
44%
learn-eu-central-1-prod-fleet01-xythos.content.blackboardcdn.com
The eigenvalue you must keep is ₁ =q₁1a +912 a j
Note that if A₁ is real valued that 912 = 0
7. The general solution of a non homogeneous state-space equation is given below. Use the initial
conditions to determine the value of C₁.
[*] = ₁²[2¹₁] + ₁₂e-²¹ [2₁] +9
1
given x(0) = 0, x'(0) = 2
You calculated that C₁ = 913+914 j
Note that if C₁ is a real number that 914 = 0.
8. The specific solution of a state space equation is given below. Express the solution Q
terms of sines and cosines using the Euler representation of the complex exponential.
[9] = 0.3 je¹ [] -0.3 je¯¹ [¹]
You calculated that Q = qiscos(t) +916sin(1)
9. Two swings are hanging from a 3 meter rectangular wood beam embedded at both ends. Swing 1 is
attached at x = 1 m and swing 2 is attached at x = 2m. The person on swing 1 applies a point
force of A Newton and the person on swing 2 applies a point force of B newton downwards.
AS
The differential equation is y' = A 8(x-1) +
) + / BE
B8(x-2) with the four boundary
conditions y(0) = 0 and y'(0) = 0 y'(0) = P and y'"(0) = Q. Determine the deflection of the
beam as a function of x in terms of A, B, E, I, P and Q using the Laplace transform. Express your
solution as a piecewise function.
9.1 Applying the Laplace transform, will result in the expression for L{y}
term 4
term 1
P
term 2
Q A
+
+
term 3
1
L{y} =
-9185
B
1
e920s
(s**) (5917) EI (s**)
EI (¹⁹)
9.2 Term 1 can be expressed as L{P 9211"}
9.3 Term 4 can be expressed as
L{B922 (1923) ¹24 H(1-**
-**)}
9.4 In expressing the solution y as a piecewise function, the part corresponding to the interval
1 < x < 2 has non zero terms:
y(x) =P**+Q**
...9250
y(x) =P**+Q**+ A
EI
... 925 = 1
... 925 = 2
y(x) =P**+Q**+
+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc83796e5-c9dd-4410-b2cd-8792a3270d87%2F29a0a2aa-f8d3-4c3e-b451-0b99676b0aeb%2Frdqdcgf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:.Telkom-Monate LTE
11:04
44%
learn-eu-central-1-prod-fleet01-xythos.content.blackboardcdn.com
The eigenvalue you must keep is ₁ =q₁1a +912 a j
Note that if A₁ is real valued that 912 = 0
7. The general solution of a non homogeneous state-space equation is given below. Use the initial
conditions to determine the value of C₁.
[*] = ₁²[2¹₁] + ₁₂e-²¹ [2₁] +9
1
given x(0) = 0, x'(0) = 2
You calculated that C₁ = 913+914 j
Note that if C₁ is a real number that 914 = 0.
8. The specific solution of a state space equation is given below. Express the solution Q
terms of sines and cosines using the Euler representation of the complex exponential.
[9] = 0.3 je¹ [] -0.3 je¯¹ [¹]
You calculated that Q = qiscos(t) +916sin(1)
9. Two swings are hanging from a 3 meter rectangular wood beam embedded at both ends. Swing 1 is
attached at x = 1 m and swing 2 is attached at x = 2m. The person on swing 1 applies a point
force of A Newton and the person on swing 2 applies a point force of B newton downwards.
AS
The differential equation is y' = A 8(x-1) +
) + / BE
B8(x-2) with the four boundary
conditions y(0) = 0 and y'(0) = 0 y'(0) = P and y'"(0) = Q. Determine the deflection of the
beam as a function of x in terms of A, B, E, I, P and Q using the Laplace transform. Express your
solution as a piecewise function.
9.1 Applying the Laplace transform, will result in the expression for L{y}
term 4
term 1
P
term 2
Q A
+
+
term 3
1
L{y} =
-9185
B
1
e920s
(s**) (5917) EI (s**)
EI (¹⁹)
9.2 Term 1 can be expressed as L{P 9211"}
9.3 Term 4 can be expressed as
L{B922 (1923) ¹24 H(1-**
-**)}
9.4 In expressing the solution y as a piecewise function, the part corresponding to the interval
1 < x < 2 has non zero terms:
y(x) =P**+Q**
...9250
y(x) =P**+Q**+ A
EI
... 925 = 1
... 925 = 2
y(x) =P**+Q**+
+
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