(Hanging cable) It can be shown that the curve y(x) of an inextensible flexible homogeneous
cable hanging between two fixed points is obtained by solving y"  k 1  y' 2 , where the
constant k depends on the weight. This curve is called a catenary (from Latin catena ＝ the
chain). Find y(x), assuming k＝1 and those fixed points are (－1, 0) and (1,0) in a vertical
xy-plane.
y"  k 1  y' 2
Set y' 
dy
dx
k  1 and y  1  0, y1  0
 z, y"  dz The original equation becomes as
dx
dz
 k 1 z2
dx

dz
1 z 2
#  k  dx
ln z  1  z 2  kx  C *
z  1  z 2  C1e kx; C1  eC *
1  z 2  C1ekx  z

2  C12e 2kx  z 2  2C1ze kz
kx 2 2kx
kx
C

2C1 ze kx  C12 e 2kx  1; z  e
C1 e
 1  1 e kx  e
2C1
2
2C1
1  z 2  C1e kx  z
Since z 
 kx 
 kx 


dy
 y   zdx  C 2  1   C1e kx  e
dx  C 2  1  C1e kx  e

 C 2
2 
C1 
2k 
C1 
dx
 kx 

General sol. y  1  C1e kx  e
  C2
2k 
C1 
Substitute the boundary values,  1, 0 and 1, 0 and k = 1 into the general solution
1  C e 1  e  C  0


2
2 1
C1 
1  C e  e 1  C  0


2
2 1
C1 
1 －○
2
○
1
○
2
○
e  e1  C1 C11   0; C1 C11  0
 e  e 1 

C1  1, C 2   
2
C1  1 or  1
 e  e 1 

C1  1, C 2  
2
Particular sol.
 e  e 1 

x
x
1
y  e e

2
2


 e  e 1 

x
x
1
o ry   e  e

2
2


depends on
the choice of y-axis orientation.
# Set z  tan  , dz  sec 2  d

dz
1 z 2

sec 2 
2
d   sec  d   sec d  ln sec   tan 
1 tan 2 
sec 2 
 ln 1  x 2  x
另解：
Set y ' 
dy
dx
dy
 z, y"  dz  dz
 z dz
dx
zdz
z dz  k 1  z 2 ;
dy

zdz
1 z2
dy dx
1 z
2
dy
 kdy Integrate both sides
 k  dy  C1   dy  C1
k 1
1  z 2 2  y  C1; 1  z 2   y  C1 2 ; z 2   y  C1 2  1
1
dy

dx
 y  C1 2 1; 
ln y  C1 
y  C1 
dy
 y  C1 2 1
  dx  C *
 y  C1 2  1  x  C *
ln y  C1 
 y  C1 2  1  x  C *
 y  C1 2  1  C2 e x ; C2  eC *
 y  C1 2  1  C2 e x   y  C1 
 y  C1 2  1  C2 e x   y  C1 
2
 C22 e 2 x   y  C1 2  2C2 e x  y  C1 


 1  C22 e 2 x  2C2 e x  y  C1   C22 e 2 x  2C2 e x y  2C1C2 e x  C2 e x C2 e x  2C1  2C2 e x y




2C 2 e x y  C 2 e x C 2 e x  2C1  1; y  1 C 2 e x  2C1  1 e  x
2
General sol.: y  1 C2 e x  1 e  x   C1
2 
C2

2C 2
With the boundary conditions  1, 0, 1, 0
We get
From
1
○
1 C e 1  1 e  C  0
1
2  2
C 2 
1
○
1 C e  1 e 1   C  0
1
2  2
C2

2
○
and

C2  1, C1   1 e  e 1
2
○
2

 
Particular sol.: y  1 e x  e  x  1 e  e 1
2
2

