equation of catenary via calculus of variations

Using the mechanical principle that the centre of mass itself as low as possible, determine the equation of the curve formed by a $l$ when supported at its ends in the points  $P_1=(x_1,y_1)$  and  $P_2=(x_2,y_2)$.

We have an isoperimetric problem

$\text{to minimise}\mathit{\hspace{1em}}{\displaystyle {\int }_{P_1}^{P_2}}y𝑑s$(1)

under the constraint

${\displaystyle {\int }_{P_1}^{P_2}}𝑑s=l,$(2)

where both the path integrals are taken along some curve $c$.  Using a Lagrange multiplier $\lambda$, the task changes to a free problem

${\displaystyle {\int }_{P_1}^{P_2}}(y-\lambda )𝑑s={\displaystyle {\int }_{x_1}^{x_2}}(y-\lambda )\sqrt{1+y^{\prime 2}}|dx|=\text{min}!$(3)

(cf. example of calculus of variations).

The Euler–Lagrange differential equation (http://planetmath.org/EulerLagrangeDifferentialEquation), the necessary condition for (3) to give an extremal $c$, reduces to the Beltrami identity

$$(y-\lambda )\sqrt{1+y^{\prime 2}}-y^{\prime }\cdot (y-\lambda )\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}\equiv \frac{y-\lambda }{\sqrt{1+y^{\prime 2}}}=a,$$

where $a$ is a constant of integration.  After solving this equation for the derivative $y^{\prime }$ and separation of variables, we get

$$\pm \frac{dy}{\sqrt{{(y-\lambda )}^2-a^2}}=\frac{dx}{a}$$

which may become clearer by notating  $y-\lambda :=u$;  then by integrating

$$\pm \frac{du}{\sqrt{u^2-a^2}}=\frac{dx}{a}$$

we choose the new constant of integration $b$ such that  $x=b$  when  $u=a$:

$$\pm {\int }_a^u\frac{du}{\sqrt{u^2-a^2}}={\int }_b^x\frac{dx}{a}$$

We can write two equivalent (http://planetmath.org/Equivalent3) results

$$\ln \frac{u+\sqrt{u^2-a^2}}{a}=+\frac{x-b}{a},\ln \frac{u-\sqrt{u^2-a^2}}{a}=-\frac{x-b}{a},$$

i.e.

$$\frac{u+\sqrt{u^2-a^2}}{a}=e^{+\frac{x-b}{a}},\frac{u-\sqrt{u^2-a^2}}{a}=e^{-\frac{x-b}{a}}.$$

Adding these allows to eliminate the square roots and to obtain

$$u=\frac{a}{2}\left(e^{\frac{x-b}{a}}+e^{-\frac{x-b}{a}}\right),$$

or

$y-\lambda =a\cosh {\displaystyle \frac{x-b}{a}}.$(4)

This is the sought form of the equation of the chain curve.  The constants $\lambda ,a,b$ can then be determined for putting the curve to pass through the given points $P_1$ and $P_2$.