Question

Change the independent variable to simplify the Euler equation, and then find a first integral of it.

$\mathbf{4}\mathbf{.}\mathbf{}\mathbf{}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{\int }}}\mathit{y}\sqrt{\mathbf{y}{\mathbf{\text{'}}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathit{d}\mathit{x}$

Short answer

Answer

The first integral of the Euler equation is $x\text{'}=\frac{C}{y\sqrt{y^2-C^2}}$.

Step-by-step solution

Given Information

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

Definition of Euler equation    

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

Find the Euler equation of the given function

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

First, let’s change the variables as follows:

$$\begin{gathered} \frac{\partial F}{\partial x\text{'}}=\frac{x\text{'}{y}^3}{1+x{\text{'}}^2{y}^2} \\ \frac{\partial F}{\partial x}=0 \end{gathered}$$

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

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

Let $F\left(x,y,y\text{'}\right)=y\sqrt{x{\text{'}}^2{y}^2+1}$.

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

Differentiate F with respect to x' and x.

$$\begin{gathered} \frac{\partial F}{\partial x\text{'}}=\frac{x\text{'}{y}^3}{\sqrt{1+x{\text{'}}^2{y}^2}} \\ \frac{\partial F}{\partial x}=0 \end{gathered}$$

The Euler becomes as shown below.

$$\begin{gathered} \frac{d}{dx}\left[\frac{x\text{'}{y}^3}{\sqrt{1+x{\text{'}}^2{y}^2}}\right]=0 \\ \frac{x\text{'}{y}^3}{\sqrt{1+x{\text{'}}^2{y}^2}}=C \end{gathered}$$

The differential equation is $x\text{'}=\frac{C}{y\sqrt{y^4-C^2}}$

Therefore, the first integral of the Euler equation is $x\text{'}=\frac{C}{y\sqrt{y^4-C^2}}$