Question

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.

7.${\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}{\mathit{e}}^{\mathbf{x}}\sqrt{\mathbf{1}\mathbf{+}\mathbf{y}{\mathbf{\text{'}}}^{\mathbf{2}}}\mathit{d}\mathit{x}$Hint: In the last integration, let$\mathit{u}\mathbf{=}{\mathit{e}}^{\mathbf{x}}$ and see Chapter 5, Problem 1.6.

Short answer

The Euler equation for the given integral ${\int }_{x_1}^{x_2}{e}^x\sqrt{1+y{\text{'}}^2}$ is$y=\arctan \left(\frac{\sqrt{e^{2x}-C^2}}{C}\right)+B.$

Step-by-step solution

Given Information.

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

Definition of Euler equations.

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.

Write and solve Euler equation.

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

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

Calculate the required derivatives.

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

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" $y\text{'}=\pm \frac{C^2}{\sqrt{e^{2x}-C^2}}$

Integrate the expression to obtain $y$.

$y=\int \frac{C^2}{\sqrt{e^{2x}-C^2}}dx$

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

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