Question

Question: The differential equation of a hanging chain supported at its ends is

${\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{2}}\mathbf{=}{\mathbf{k}}^{\mathbf{2}}\left(1+y^{\text{'}2}\right)$. Solve the equation to find the shape of the chain.

Short answer

The solutions of the differential equation$y=\frac{1}{\pm k}\cosh (\pm kx+A)+B$

Step-by-step solution

Given information from question

The differential equation is $y^{\text{'}\text{'}2}=k^2\left(1+y^{\text{'}2}\right)$.

Differential equation

A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of a function at a given moment define its rate of change.

Exploit the chain rule for derivatives

The given equation is lacking by independent variable, thus it can be significantly simplified by using the substitution, $y^{\text{'}}\equiv p$also implying by $y^{\text{'}\text{'}}=p^{\text{'}}$exploiting the chain rule for derivatives. Notice that the equation lacking the variable x as well, but that this transformation leads to significantly simpler solution.

Apply the square root operation on the equation
$y^{\text{'}\text{'}2}=k^2\left(1+y^{\text{'}2}\right)\Rightarrow y^{\text{'}\text{'}}=\pm k\sqrt{1+y^2}$
Notice the sign since the signature of the constant is not given.
$p^{\text{'}}=\pm k\sqrt{1+p^2}$
And since the previous equation is manifestly separable, so by separating it to obtain:
$$\begin{gathered} \frac{dp}{dx}=\pm k\sqrt{1+p^2} \\ \Rightarrow \frac{dp}{\sqrt{1+p^2}}=\pm kdx \end{gathered}$$
Bearing in mind ${\cosh }^{2}x={\sinh }^{2}x+1$, and substitute the $p=\sinh u\Rightarrow dp=\cosh udu$
$$\begin{gathered} \frac{dp}{\sqrt{1+p^2}}=\pm kdx \\ \frac{\cosh udu}{\sqrt{1+\sinh u^2}}=\pm kdx \\ \frac{\cosh udu}{\cosh u}=\pm kdx \end{gathered}$$

Integrate and use formula ∫dx=x+C

By integration and using formula $\int dx=x+C$then the previous equation

$$\begin{gathered} u=\pm kx+A \\ \Rightarrow {\sinh }^{-1}p=\pm kx+A \\ \Rightarrow p=\sinh (\pm kx+A) \end{gathered}$$
Since$p=\frac{dy}{dx}$ so they can integrate once again and use the table integral
$$\begin{gathered} \int \sinh (ax+b)dx=\frac{1}{a}\cosh (ax+b)+C \\ y=\frac{1}{\pm k}\cosh (\pm kx+A)+B \end{gathered}$$
Thus, the solutions of the differential equation$y=\frac{1}{\pm k}\cosh (\pm kx+A)+B$