91Ó°ÊÓ

The given integral is ${∫}_{x_1}^{x_2}{e}^x\sqrt{1+y{\text{'}}^2}ds$.

The Euler–Lagrange equations are a series of second-order ordinary differential equations whose solutions are stationary points of the specified action functional in the calculus of variations and classical mechanics.

Let$F=e^x\sqrt{1+y{\text{'}}^2}$

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

Calculate the required derivatives.

$$\begin{gathered} \frac{∂F}{∂y\text{'}}=e^x\frac{y\text{'}}{\sqrt{1+y{\text{'}}^2}} \\ \frac{∂F}{∂y}=0 \end{gathered}$$

Further, there is no need to calculate the derivative with respect to$x$because it is zero in the context of the Euler equation and therefore the whole expression is constant.

$$\begin{gathered} \frac{d}{dx}\left[\frac{e^{x}y\text{'}}{\sqrt{1+y{\text{'}}^2}}\right]=0 \\ \frac{e^{x}y\text{'}}{\sqrt{1+y{\text{'}}^2}}=C \end{gathered}$$

Solve for$y\text{'}$. Square both sides of the equation and multiply by denominator to obtain:

$$\begin{gathered} e^{2x}y{\text{'}}^2=C^2\left(1+y{\text{'}}^2\right) \\ \left(e^{2x}-C^2\right)y{\text{'}}^2=C^2 \\ y{\text{'}}^2=\frac{C^2}{e^{2x}-C^2} \end{gathered}$$

Therefore,localid="1665121441014" $y\text{'}=Â\pm \frac{C^2}{\sqrt{e^{2x}-C^2}}$

Integrate the expression to obtain $y$.

$y=∫\frac{C^2}{\sqrt{e^{2x}-C^2}}dx$

localid="1665121337150" $y=\arctan \left(\frac{\sqrt{e^{2x}-C^2}}{C}\right)+B$

Therefore, the Euler equation is $y=\arctan \left(\frac{\sqrt{e^{2x}-C^2}}{C}\right)+B$.