28 Chapter 0 Ordinary Differential Equations
Substituting the derivative for the slope and making some algebraic adjust-
ments, we obtain
T(
u′(x+ x)−u′(x))
=f(x) x.Dividing through by xyields
Tu′(x+ x)−u′(x)
x =f(x).In the limit, as xapproaches 0, the difference quotient in the left member
becomes the second derivative ofu, and the result is the equation
Td(^2) u
dx^2
=f(x), (3)
which is valid forxin the range 0<x<a, where the cable is located. In addi-
tion,u(x)must satisfy the boundary conditions
u( 0 )=h 0 , u(a)=h 1. (4)
For any particular case, we must choose an appropriate model for the load-
ing,f(x). One possibility is that the cable is hanging under its own weight of
wunits of weight per unit length of cable. Then in Eq. (2), we should put
f(x) x=w xs x,
wheresrepresents arc length along the cable. In the limit, as xapproaches 0,
s/ xhas the limit
limx→ 0 s
x
=
√
1 +
(du
dx) 2
.
Therefore, with this assumption, the boundary value problem that determines
the shape of the cable is
d^2 u
dx^2=w
T√
1 +
(du
dx) 2
, 0 <x<a, (5)u( 0 )=h 0 , u(a)=h 1. (6)Notice that the differential equation is nonlinear. Nevertheless, we can find its
general solution in closed form and satisfy the boundary conditions by appro-
priate choice of the arbitrary constants that appear. (See Exercises 4 and 5.)
Another case arises when the cable supports a load uniformly distributed in
the horizontal direction, as given by
f(x) x=w x.