Question

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

$\mathbf{2}\mathbf{.}\mathbf{}\mathbf{}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\frac{\sqrt{\mathbf{1}\mathbf{+}\mathbf{y}{\mathbf{\text{'}}}^{\mathbf{2}}}}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{dx}$

Short answer

The first integral of the Euler equation is$\mathrm{x}\text{'}=\frac{{\mathrm{Cy}}^2}{\sqrt{1-{\mathrm{C}}^2{\mathrm{y}}^4}}$

Step-by-step solution

Given Information

The given integral is${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\frac{\sqrt{1+\mathrm{y}{\text{'}}^2}}{{\mathrm{y}}^2}\mathrm{dx}$

Definition of Euler equation

For the integral, $\mathrm{I}\left(\mathrm{\epsilon }\right)={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{F}\left(\mathrm{x},\mathrm{y},\mathrm{y}\text{'}\right)\mathrm{dx}$the Euler equation defined as$\frac{\mathrm{d}}{\mathrm{dy}}\frac{\partial \mathrm{F}}{\partial \mathrm{x}\text{'}}-\frac{\partial \mathrm{F}}{\partial \mathrm{x}}=0$

Find Euler equation of the given function

The given integral is ${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\frac{\sqrt{1+\mathrm{y}{\text{'}}^2}}{{\mathrm{y}}^2}\mathrm{dx}$

First let’s change the variables as follows

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

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

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

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

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

Differentiate$\mathrm{F}$with respect to$\mathrm{x}\text{'}$and$\mathrm{x}$

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

The Euler become

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dy}}\left[\frac{1}{{\mathrm{y}}^2\sqrt{1+\mathrm{x}{\text{'}}^2}}\right]=0 \\ \frac{1}{{\mathrm{y}}^2\sqrt{1+\mathrm{x}{\text{'}}^2}}=\mathrm{C} \\ 1+\mathrm{x}{\text{'}}^2=\frac{1}{{\mathrm{C}}^2{\mathrm{y}}^4} \end{gathered}$$

Therefore, the differential equation is$\mathrm{x}\text{'}=\pm \frac{{\mathrm{Cy}}^2}{\sqrt{1-{\mathrm{C}}^2{\mathrm{y}}^4}}$

Since minus can be absorbed into the constant.

$\mathrm{x}\text{'}=\frac{{\mathrm{Cy}}^2}{\sqrt{1-{\mathrm{C}}^2{\mathrm{y}}^4}}$

Therefore, the first integral of the Euler equation is$\mathrm{x}\text{'}=\frac{{\mathrm{Cy}}^2}{\sqrt{1-{\mathrm{C}}^2{\mathrm{y}}^4}}$