91Ó°ÊÓ

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

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

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

Now calculate the required derivatives.

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

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.

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

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

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

Therefore,

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

Integrate the expression to obtain $B$.

$$\begin{gathered} y=∫\frac{C^2}{\sqrt{x-C^2}}dx \\ y=2C\sqrt{x-C^2}+B \end{gathered}$$

Let’s rewrite the expression in a simple way by moving $B$to the left side and then square both the sides.

$$\begin{gathered} y-B=2C\sqrt{x-C^2} \\ {\left(y-B\right)}^2=4C^2\left(x-C^2\right) \end{gathered}$$which corresponds to a parabola.

Therefore, the Euler equation is ${\left(y-B\right)}^2=4C^2\left(x-C^2\right)$.