Solving a second Order ODE of a catenary curve
Mia Lopez 
Given the second order ODE of a catenary curve $$y''=a\sqrt{1+(y')^2}$$
With initial conditions $y(0)=0$ and $y'(0)=0$
The equation was first rewritten as $2y''y'''=2a^{2}y'y''$ and then dividing both sides by $$2y''$$ and letting $z=y'$ we would get the equation $$z''=a^2z$$
intital conditions $z(0)=0$ and $z'(0)=a$
I came across this question while i was reading the following article. I couldn't fully understand what the article meant such as how did they form the Differential equation of the Caterny curve and how did they managed to solve the differential equation. Could anyone Please explain. Thanks
$\endgroup$ 61 Answer
$\begingroup$I'm starting from the "Since the sum of the forces..", since that is where the confusion starts.
Notice that the tangential force is the gradient of the Catenary (curve) and the slope can be found by using
$$\frac{\text{rise}}{\text{run}} = \frac{W}{H} = \frac{w}{h} \ \ \ (1)$$
Where $w$ and $h$ are the magnitudes of $W$ and $H$, respectively. But the paper also states that "the magnitude, $w$, is proportional to the length $s$ of the chain between the origin and the point $(x, y)$" i.e.
$$w \propto s \implies \frac{w}{h} = \frac{\mu s}{h} \ \ \ (2)$$
where $\mu$ is the weight density of the chain.
So we have the gradient of the Catenary, $y'(x)$, is given by
$$\begin{align} y'(x) &= \frac{\text{rise}}{\text{run}} \\ &= \frac{\mu s}{h} \ \ \ (3)\\ \end{align}$$
But remember, $s$ is the length of the chain from the point $(x, y)$ to the origin, also known as the arc length, which has an integral equation
$$s = \int_{0}^{x} \sqrt{1 + y'(t)^{2}} dt \ \ \ (4)$$
Substituting $(4)$ into $(3)$ gives
$$y'(x) = \frac{\mu}{h} \int_{0}^{x} \sqrt{1 + y'(t)^{2}} dt$$
Taking the derivative of both sides with respect to $x$ (the integral is just an application of the fundamental theorem of calculus) and setting $\alpha = \frac{\mu}{h}$, we get
$$y''(x) = \alpha \sqrt{1 + y'(x)^{2}}$$
They then solved the equation by squaring both sides of the equation and taking the derivative of both sides with respect to $x$, using implicit differentiation i.e.
$$\begin{align} y''^{2} &= \alpha^{2} (1 + y'^{2}) \\ \implies \frac{d}{dx} y''^{2} &= \frac{d}{dx} \alpha^{2} (1 + y'^{2}) \\ \implies 2y'' y''' &= 2 \alpha^{2} y' y'' \\ \end{align}$$
Dividing by $2y''$
$$\implies y''' = \alpha^{2} y' \ \ \ (5)$$
Then they used the change of variables,
$$z = y' \implies z' = y'' \implies z'' = y'''$$
Which makes $(5)$ become
$$z'' = \alpha^{2}z$$
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