JA 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 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) d²y dz? y(0) = The solution of this problem is y = y(x), where y(x) =
JA 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 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) d²y dz? y(0) = The solution of this problem is y = y(x), where y(x) =
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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Transcribed Image Text:JA 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
dz?
y(0)
The solution of this problem is y = y(x), where
y(x) =
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