A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the density p = 1.5 Ib/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P and the x-axis is a = 2 ft below P1. If the absolute value of tension force in the cable at point P is T = 2, then the initial value problem that governs the shape of the cable is (Use the notation dy/d for the first derivative) d'y 0.75 sqrt [1+(dy/dx)^2] zp y(0) = 2 2 (0) = The solution of this problem is y = y(x), where y(x) = 4/3cosh (0.75)x+1/2
A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the density p = 1.5 Ib/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P and the x-axis is a = 2 ft below P1. If the absolute value of tension force in the cable at point P is T = 2, then the initial value problem that governs the shape of the cable is (Use the notation dy/d for the first derivative) d'y 0.75 sqrt [1+(dy/dx)^2] zp y(0) = 2 2 (0) = The solution of this problem is y = y(x), where y(x) = 4/3cosh (0.75)x+1/2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please show all the steps AND make sure the last one's answer is correct
![A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the
density p = 1.5 lb/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P and the x-axis is a = 2 ft below P. If the absolute value of tension
force in the cable at point P is T = 2, then the initial value problem that governs the shape of the cable is (Use the notation dy/dx for the first derivative)
d'y
0.75 sqrt [1+(dy/dx)^2]
dz?
y(0)
2
(0)
The solution of this problem is y = y(x), where
y(x) = 4/3cosh (0.75)x+1/2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb6da5ff5-4f9e-46f1-b6af-425e0be2530f%2Fee2773c8-26da-4842-9f3b-908a1e508c8f%2Fuosru5a_processed.png&w=3840&q=75)
Transcribed Image Text:A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the
density p = 1.5 lb/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P and the x-axis is a = 2 ft below P. If the absolute value of tension
force in the cable at point P is T = 2, then the initial value problem that governs the shape of the cable is (Use the notation dy/dx for the first derivative)
d'y
0.75 sqrt [1+(dy/dx)^2]
dz?
y(0)
2
(0)
The solution of this problem is y = y(x), where
y(x) = 4/3cosh (0.75)x+1/2
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