Write and solve the Euler equations to make the following integrals stationary. In solving the Euler equations, the integrals in Chapter 5, Section 1, may be useful.

5.${\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\left(y{\text{'}}^2+y^2\right)\mathit{d}\mathit{s}$

The given integral is ${\int }_{x_1}^{x_2}\left(y{\text{'}}^2+y^2\right)dx$.

The solutions of the Euler-Lagrange equations, which are stationary points of the defined action functional in the calculus of variations and classical mechanics, are a set of second-order ordinary differential equations.

Let $F=y^2+y{\text{'}}^2$.

Write the Euler equation as $\frac{d}{dx}\frac{\partial F}{\partial y\text{'}}-\frac{\partial F}{\partial y}=0$.

Now, calculate the required derivatives.

$$\begin{gathered} \frac{\partial F}{\partial y\text{'}}=2y\text{'} \\ \frac{d}{dx}\frac{\partial F}{\partial y\text{'}}=\frac{d}{dx}\left[2y\text{'}\right] \\ =2y\text{'}\text{'} \\ \frac{\partial F}{\partial y}=2y \end{gathered}$$

Therefore, the Euler equation reads:

$$\begin{gathered} 2y\text{'}\text{'}-2y=0 \\ y\text{'}\text{'}-y=0 \end{gathered}$$

Now insert second-order linear differential equation $y=A{e}^{\alpha x}$into the given differential equation.

$$\begin{gathered} y\text{'}\text{'}-y=0 \\ A{\alpha }^2{e}^{\alpha x}-A{e}^{\alpha x}=0 \\ {\alpha }^2=1 \\ \alpha =\pm 1 \end{gathered}$$

The solution to the second-order linear differential equation has two solutions, so the general solution is their linear combination for two different values of $\alpha$.

$y=A{e}^x+B{e}^{-x}$

Therefore, the Euler equation is $y=A{e}^x+B{e}^{-x}$.