91Ó°ÊÓ

The given integral is $\underset{x_1}{\overset{x_2}{∫}}\sqrt{1+y^{2}y{\text{'}}^2}dx$

For the integral $\mathit{I}\left(\mathrm{Î\mu }\right)\mathbf{=}\mathit{F}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{∫}}}\left(x,y,y\text{'}\right)\mathit{d}\mathit{x}$,the Euler equation is mathematically presented as $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\frac{\mathbf{d}\mathbf{F}}{\mathbf{d}\mathbf{y}\mathbf{\text{'}}}\mathbf{-}\frac{\mathbf{d}\mathbf{F}}{\mathbf{d}\mathbf{y}}\mathbf{=}\mathbf{0}$.

The given integral is $\underset{x_1}{\overset{x_2}{∫}}\sqrt{1+y^{2}y{\text{'}}^2}dx$.

First let’s change the variables as follows:

$$\begin{gathered} dx=x\text{'}dy \\ y\text{'}=\frac{1}{x\text{'}} \end{gathered}$$

By using the two equations above, we can rewrite the integral.

$$\begin{gathered} ∫\sqrt{1+y^{2}y{\text{'}}^2}dx=∫\sqrt{1+y^{2}y{\text{'}}^2}dx \\ =∫\sqrt{1+y^{2}y{\text{'}}^{-2}}dx \\ =∫\sqrt{1+y{\text{'}}^2{y}^2}dx \end{gathered}$$

Let $F\left(x,y,y\text{'}\right)=\sqrt{x{\text{'}}^2+y^2}$

By the definition of the Euler equation,$\frac{d}{dy}\frac{∂F}{∂x\text{'}}-\frac{∂F}{∂x}=0$.

Differentiate F with respect to x' and x.

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

The Euler becomes as shown below.

role="math" localid="1655312972293" $$\begin{gathered} \frac{d}{dy}\left[\frac{x\text{'}}{\sqrt{y^2+x{\text{'}}^2}}\right]=0 \\ \frac{x\text{'}}{\sqrt{y^2+x{\text{'}}^2}}=C \\ x{\text{'}}^2=\frac{C^2{y}^2}{1-C^2} \end{gathered}$$

Therefore, the differential equation is $x{\text{'}}^2=\frac{C^2{y}^2}{1-C^2}$.

The differential equation is

Integrate with respect to.

Here,is the integration constant. By redefining the constant, the final result takes a much simpler form.