Question

Write and solve the Euler equations to make the following integrals stationary. In solving the Euler equations, the integrals in Chapter 5, Section 1, may be useful.

${\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\left(y{\text{'}}^2+\sqrt{y}\right)\mathit{d}\mathit{x}$

Short answer

After solving the Euler equations to make the integral stationary, the answer is ${\left(x+k\right)}^2=\frac{16}{9}\left(\sqrt{y}+c\right){\left(\sqrt{y}-2c\right)}^2$.

Step-by-step solution

Given Information.

The given integral is $I={\int }_{x_1}^{x_2}\left(y{\text{'}}^2+\sqrt{y}\right)dx$.

Definition/ Concept.

The formulas related to this questions are $\mathbf{\int }\frac{\mathbf{d}\mathbf{x}}{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{a}}}\mathbf{=}{\mathbf{sinh}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{x}}{\mathbf{a}}$, $\mathbf{\int }\frac{\mathbf{d}\mathbf{y}}{\sqrt{{\mathbf{y}}^{\mathbf{2}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}}}\mathbf{=}{\mathbf{cosh}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{y}}{\mathbf{a}}$etc.

Solve the Euler equation to make integral stationary.

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

Then:

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

By Euler’s equation $\frac{d}{dx}\left(\frac{\partial F}{\partial y\text{'}}\right)-\frac{\partial F}{\partial y}=0$, we have:

$$\begin{gathered} \frac{d}{dx}\left(\frac{\partial F}{\partial y\text{'}}\right)-\frac{\partial F}{\partial y}=0 \\ \frac{d}{dx}\left(2y\text{'}\right)-\frac{1}{2\sqrt{y}}=0 \\ 2y\text{'}\text{'}-\frac{1}{2\sqrt{y}}=0 \end{gathered}$$

Let $y\text{'}=p.\mathrm{So},$

$$\begin{gathered} y\text{'}\text{'}=\frac{dp}{dx} \\ y\text{'}\text{'}=\frac{dp}{dy}.\frac{dy}{dx} \\ y\text{'}\text{'}=p\frac{dp}{dy} \end{gathered}$$

So:

$$\begin{gathered} 2p\frac{dp}{dy}-\frac{1}{2\sqrt{y}}=0 \\ 2pdp-\frac{dy}{2\sqrt{y}}=0 \end{gathered}$$

Now integrating the above equation as:

$$\begin{gathered} p^2-\frac{1}{2}\frac{y^{-{\displaystyle \frac{1}{2}}+1}}{2\left({\displaystyle \frac{1}{2}}\right)}=c \\ {p}^2=c+\sqrt{y} \\ p=\sqrt{c+\sqrt{y}} \\ \frac{dy}{dx}=\sqrt{c+\sqrt{y}} \end{gathered}$$

Now by the variable separable DE, we can write:

$$\begin{gathered} \frac{dy}{dx}=\sqrt{c+\sqrt{y}} \\ \int \frac{dy}{\sqrt{c+\sqrt{y}}}=\int dx+k \end{gathered}$$

Let $y=t^2,dy=2tdt$, we get:

$$\begin{gathered} x+k=2\int \frac{t}{\sqrt{c+t}}dt \\ x+k=2\int \frac{t+c-c}{\sqrt{c+t}}dt \\ x+k=2\left[\int \sqrt{c+}tdt-\int \frac{c}{\sqrt{c+t}}dt\right] \\ x+k=2\times \frac{{\left(c+t\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}-2c\times \frac{{\left(c+t\right)}^{-{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}} \end{gathered}$$

Solving further as:

$$\begin{gathered} x+k=\frac{4}{3}{\left(c+t\right)}^{\frac{3}{2}}-4c{\left(c+t\right)}^{\frac{1}{2}} \\ x+k=\frac{4}{3}{\left(c+t\right)}^{\frac{1}{2}}\left(t-2c\right) \\ {\left(x+k\right)}^2=\frac{16}{9}\left(\sqrt{y}+c\right){\left(\sqrt{y}-2c\right)}^2 \end{gathered}$$

Therefore, the answer is ${\left(x+k\right)}^2=\frac{16}{9}\left(\sqrt{y}+c\right){\left(\sqrt{y}-2c\right)}^2$.