Editing Euler–Lagrange equation (section)

==Example==
A standard example{{Citation needed|reason=this example seems like a poor one as it is not even presented as a dynamical system|date=July 2023}} is finding the real-valued function ''y''(''x'') on the interval [''a'', ''b''], such that ''y''(''a'') = ''c'' and ''y''(''b'') = ''d'', for which the [[path (topology)|path]]  [[arc length|length]] along the [[Curve#length of a curve|curve]] traced by ''y'' is as short as possible. 
:<math> \text{s} = \int_{a}^{b} \sqrt{\mathrm{d}x^2+\mathrm{d}y^2} = \int_{a}^{b} \sqrt{1+y'^2}\,\mathrm{d}x,</math>
the integrand function being <math display="inline"> L(x,y, y') = \sqrt{1+y'^2} </math>.

The partial derivatives of ''L'' are:
:<math>\frac{\partial L(x, y, y')}{\partial y'} = \frac{y'}{\sqrt{1 + y'^2}} \quad \text{and} \quad
\frac{\partial L(x, y, y')}{\partial y} = 0.</math>
By substituting these into the Euler–Lagrange equation, we obtain
:<math> 
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} \frac{y'(x)}{\sqrt{1 + (y'(x))^2}} &= 0 \\ 
\frac{y'(x)}{\sqrt{1 + (y'(x))^2}} &= C = \text{constant} \\
\Rightarrow y'(x)&= \frac{C}{\sqrt{1-C^2}} =: A \\
\Rightarrow y(x) &= Ax + B
\end{align}
</math>
that is, the function must have a constant first derivative, and thus its [[graph of a function|graph]] is a [[straight line]].