Question

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

$\mathbf{3}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{\int }}}\frac{{\mathbf{x}}^{\mathbf{t}\mathbf{2}}}{\sqrt{{\mathbf{x}}^{\mathbf{t}\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}\mathit{d}\mathit{y}$

Short answer

Answer

The equation $x^{4}y{\text{'}}^2=C^2{\left(1+x^{2}y{\text{'}}^2\right)}^3$can be left in this form, since the solution is not trivial and there is no need to solve it.

Step-by-step solution

Given Information

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

Definition of Euler equation

For the integral$\mathit{l}\left(\epsilon \right)\mathbf{=}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{\int }}}\left(x,y,y\right)\mathit{d}\mathit{x}$, 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 $x^{4}y{\text{'}}^2=C^2{\left(1+x^{2}y{\text{'}}^2\right)}^3$.

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.

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

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

By the definition of the Euler equation,$\frac{d}{dx}\frac{\partial F}{\partial F}-\frac{\partial F}{\partial y}=0$.

Differentiate F with respect to y' and y.

$$\begin{gathered} \frac{\partial F}{\partial y\text{'}}=\frac{y\text{'}{x}^2}{{\left(1+x^{2}y{\text{'}}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \frac{\partial F}{\partial y}=0 \end{gathered}$$

The Euler equations becomes:

$$\begin{gathered} \frac{d}{dx}\left[\frac{y\text{'}{x}^2}{{\left(1+x^{2}y{\text{'}}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right]=0 \\ \frac{y\text{'}{x}^2}{{\left(1+x^{2}y{\text{'}}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=C \\ {x}^{4}y{\text{'}}^3{\left(1+x^{2}y{\text{'}}^2\right)}^3 \end{gathered}$$

We shall leave the equation$x^{4}y{\text{'}}^3{\left(1+x^{2}y{\text{'}}^2\right)}^3$in this form, since the solution is not trivial and there is no need to solve it.