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.

3.${\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathit{x}\sqrt{\mathbf{1}\mathbf{-}\mathbf{y}{\mathbf{\text{'}}}^{\mathbf{2}}}\mathit{d}\mathit{x}$

Thegiven integral is ${\int }_{x_1}^{x_2}x\sqrt{1-y{\text{'}}^2}dx$.

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 $F=x\sqrt{1-y{\text{'}}^2}$

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

Calculate the required derivatives.

$$\begin{gathered} \frac{\partial F}{\partial y\text{'}}=\frac{xy\text{'}}{\sqrt{1-y{\text{'}}^2}} \\ \frac{\partial F}{\partial y}=0 \end{gathered}$$

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.

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

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

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

Therefore,

$y\text{'}=\pm \frac{C}{\sqrt{x^2+C^2}}$

Integrate the expression to obtain$\mathrm{y}$.

$$\begin{gathered} y=\int \frac{C}{\sqrt{x^2+C^2}}dx \\ y=C\mathrm{In}\left(\sqrt{{\mathrm{x}}^2+{\mathrm{C}}^2}+\mathrm{x}\right)+\mathrm{B} \end{gathered}$$

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