We consider the special case when the function \(L\) does not depend explicitly on \(x\) (that is, it still depends explicitly on the functions \(y(x)\) and \(y'(x)\)):
\begin{align}
L&=L(y,y')\\
\frac{\partial L}{\partial x}&=0
\tag{2.51}\end{align}
Consider the following identity:
\begin{align}
\frac{d}{dx}\left(y'\frac{\partial L}{\partial y'} -L \right)&=y''\frac{\partial L}{\partial y'}+y' \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)- \frac{dL}{dx}\\
&=y''\frac{\partial L}{\partial y'}+y' \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)-\left(\frac{\partial L}{\partial y}\frac{dy}{dx}+\frac{\partial L}{\partial y'}\frac{dy'}{dx} +\frac{\partial L}{\partial x} \right)\\
&=y''\frac{\partial L}{\partial y'}+y' \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)-\left(\frac{\partial L}{\partial y}y'+\frac{\partial L}{\partial y'}y'' \right)\\
&=y'\left[ \frac{d}{dx}\left(\frac{\partial L}{\partial y'}\right)-\frac{\partial L}{\partial y}\right]
\tag{2.52}\end{align}
where we have used the fact that:
\begin{align}
dL=\frac{\partial L}{\partial y}dy+\frac{\partial L}{\partial y'}dy' +\frac{\partial L}{\partial x}dx
\tag{2.53}\end{align}
where the last term is zero.
If the function \(L\) satisfies the Euler-Lagrange equations, this implies that:
\begin{align}
\frac{d}{dx}\left(y'\frac{\partial L}{\partial y'} -L \right)&=0\\
\therefore y'\frac{\partial L}{\partial y'} -L =k
\tag{2.54}\end{align}
where \(k\) is a constant. This is equivalent to the Euler-Lagrange equations for the case where \(L\) does not explicitly depend on \(x\). It is an easier equation to solve for \(y(x)\), since it only contains first order derivatives.
The case when there are multiple functions
Suppose that the function \(L\) depends on multiple functions \(y_1(x),\dots, y_n(x)\) and their derivatives, \(y'_1(x),\dots, y'_n(x)\), \(L=L(y_1(x),\dots, y_n(x),y'_1(x),\dots, y'_n(x))\). If \(L\) does not explicitly depend on \(x\), we have:
\begin{align}
\frac{dL}{dx}&=\frac{\partial L}{\partial y_1}y'_1+\dots +\frac{\partial L}{\partial y_n}y'_n+
\frac{\partial L}{\partial y'_1}y''_1+\dots +\frac{\partial L}{\partial y'_n}y''_n\\
&=\sum_{i=1}^n\left(\frac{\partial L}{\partial y_i}y'_i+\frac{\partial L}{\partial y'_i}y''_i \right)
\tag{2.55}\end{align}
We thus consider the identity:
\begin{align}
\frac{d}{dx}\left(\sum_{i=1}^n\left( y_i'\frac{\partial L}{\partial y_i'}\right) -L \right)&=\sum_{i=1}^n\left( y_i''\frac{\partial L}{\partial y_i'}+y_i'\frac{d}{dx}\frac{\partial L}{\partial y_i'}\right)-\frac{dL}{dx}\\
&=\sum_{i=1}^n\left( y_i''\frac{\partial L}{\partial y_i'}+y_i'\frac{d}{dx}\frac{\partial L}{\partial y_i'}\right)-\sum_{i=1}^n\left(\frac{\partial L}{\partial y_i}y'_i+\frac{\partial L}{\partial y'_n}y''_n \right)\\
&=\sum_{i=1}^n\left(y'_i\left[\frac{d}{dx}\frac{\partial L}{\partial y_i'}- \frac{\partial L}{\partial y_i} \right] \right)
\tag{2.56}\end{align}
Again, the term on the right is zero since the Euler-Lagrange equations are satisfied by each \(y_i(x)\). We thus obtain the equivalent result as we did with a single function (noting that we have a sum on the left):
\begin{align}
\frac{\partial L}{\partial x}=0\\
\therefore \sum_{i=1}^n\left( y_i'\frac{\partial L}{\partial y_i'}\right) -L=k
\tag{2.57}\end{align}
where \(k\) is a constant.