29.3. SEPARABLE DIFFERENTIAL EQUATIONS, STABILITY 549

the chain at t and at 0 respectively, as shown. Let the bottom of the chain be at the originas shown. If this chain does not move, then all these forces acting on it must balance. Inparticular,

T (x)sinθ = l (x)ρg, T (x)cosθ = T0.

Therefore, dividing these yields

sinθ

cosθ= l (x)

≡c︷ ︸︸ ︷ρg/T0.

Now letting y(x) denote the y coordinate of the hanging chain corresponding to x,

sinθ

cosθ= tanθ = y′ (x) .

Therefore, this yieldsy′ (x) = cl (x) .

Now differentiating both sides of the differential equation,

y′′ (x) = cl′ (x) = c√

1+ y′ (x)2

and soy′′ (x)√

1+ y′ (x)2= c.

Let z(x) = y′ (x) so the above differential equation becomes

z′ (x)√1+ z2

= c.

Therefore,∫ z′(x)√

1+z2dx = cx+d. Change the variable in the antiderivative letting u = z(x)

and this yields∫ z′ (x)√1+ z2

dx =∫ du√

1+u2= sinh−1 (u)+C = sinh−1 (z(x))+C.

Therefore, combining the constants of integration,

sinh−1 (y′ (x))= cx+d

and soy′ (x) = sinh(cx+d) .

Therefore,

y(x) =1c

cosh(cx+d)+ k

where d and k are some constants and c = ρg/T0. Curves of this sort are called catenaries.Note these curves result from an assumption that the only forces acting on the chain are asshown.