91Ó°ÊÓ

The given function is$F\left(x,y,y\text{'}\right)={\left(1+yy\text{'}\right)}^2$

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

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

By the definition of Euler equation $\frac{d}{dx}\frac{dF}{dy\text{'}}-\frac{dF}{dy}=0$

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

$$\begin{gathered} \frac{dF}{dy\text{'}}=2y\left(1+yy\text{'}\right) \\ \frac{d}{dx}\frac{dF}{dy\text{'}}=\frac{d}{dx}\left[2y\left(1+yy\text{'}\right)\right] \\ =2y\text{'}\left(1+yy\text{'}\right)+2y\left(y{\text{'}}^2+yy\text{'}\right) \\ \frac{dF}{dy}=2y\text{'}\left(1+yy\text{'}\right) \end{gathered}$$

The Euler become

$$\begin{gathered} 2y\text{'}\left(1+yy\text{'}\right)+2y\left(y{\text{'}}^2+yy\text{'}\text{'}\right)-2y\text{'}\left(1+yy\text{'}\right)=0 \\ 2y\left(y{\text{'}}^2+yy\text{'}\text{'}\right)=0 \end{gathered}$$

First possibility is for that$y=0$but that is a trivial solution which we shall discard.

Therefore, $y{\text{'}}^2+yy\text{'}\text{'}=0$

Now put$v\left(y\right)=\frac{dy\left(x\right)}{dx}$

role="math" localid="1665118841445" $$\begin{gathered} \frac{d^{2}y\left(x\right)}{d{x}^2}=\frac{d}{dx}\left(\frac{dy\left(x\right)}{dx}\right) \\ =\frac{d}{dx}\left(v\left(y\right)\right) \\ =\frac{dv\left(y\right)}{dy}\frac{dy}{dx} \\ =v\left(y\right)\frac{dv\left(y\right)}{dy} \end{gathered}$$

Rewrite the given differential equation using $v\left(y\right)$

$$\begin{gathered} v{\left(y\right)}^2+y\frac{dv\left(y\right)}{dy}v\left(y\right)=0 \\ v\left(y\right)\left(v\left(y\right)+y\frac{dv\left(y\right)}{dy}\right)=0 \end{gathered}$$

Again, there are two possibilities. Let us first examine

$$\begin{gathered} v\left(y\right)=0 \\ \frac{dy\left(x\right)}{dx}=0 \\ y=C_1 \end{gathered}$$

This is a trivial solution.

Now, $v\left(y\right)+y\frac{dv\left(y\right)}{dy}=0$

$$\begin{gathered} ∫\frac{dv}{v}=-∫\frac{dy}{y} \\ v\left(y\right)=\frac{C_2}{y} \end{gathered}$$

Rewrite the equation using $v\left(y\right)=\frac{dy\left(x\right)}{dx}$

$\frac{dy\left(x\right)}{dx}=\frac{C_2}{y}$

Integrating with respect to $x$

$\frac{y^2}{2}=C_{2}x+C_3$

Therefore, the curve obtained by the Euler equations is$x=a{y}^2+b$