91Ó°ÊÓ

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

For the integral$\mathit{I}\left(\mathrm{Î\mu }\right)\mathbf{=}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{∫}}}\mathit{F}\left(x,y,y\text{'}\right)\mathit{d}\mathit{x}$, 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 $\underset{x_1}{\overset{x_2}{∫}}\frac{yy{\text{'}}^2}{1+yy\text{'}}dx$.

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.

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

Let $F\left(x,y,y\text{'}\right)=\frac{y}{x\text{'}+y}$.

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.

$$\begin{gathered} \frac{∂F}{∂x\text{'}}=-\frac{y}{{\left(x\text{'}+y\right)}^2} \\ \frac{∂F}{∂x}=0 \end{gathered}$$

The Euler becomes as shown below.

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

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

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

Integrate with respect to.

$$\begin{gathered} x=∫\left(\sqrt{\frac{y}{C}}-y\right)dy \\ x=\frac{2}{3\sqrt{C}}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{2}{y}^2+A \end{gathered}$$

Here, A is the integration constant.