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Chapter 9: Q4P (page 481)

Change the independent variable to simplify the Euler equation, and then find a first integral of it.
$4 . {鈭�}_{x_{1}}^{x_{2}} y \sqrt{y'^{2} + y^{2}} d x$

Short Answer

Expert verified

Answer

The first integral of the Euler equation is $x' = \frac{C}{y \sqrt{y^{2} - C^{2}}}$ .

Step by step solution

01

Given Information

The given integral is ${鈭�}_{x_{1}}^{x_{2}} y \sqrt{y'^{2} + y^{2}} d x$ .

02

Definition of Euler equation

For the integral $I (蔚) = F {鈭�}_{x_{1}}^{x_{2}} (x, y, y') d x$ , the Euler equation is mathematically presented as $\frac{d}{d x} \frac{d F}{d y'} - \frac{d F}{d y} = 0$ .

03

Find the Euler equation of the given function

The given integral is ${鈭�}_{x_{1}}^{x_{2}} y \sqrt{y'^{2} + y^{2}} d x$ .

First, let鈥檚 change the variables as follows:

$\frac{鈭� F}{鈭� x'} = \frac{x' y^{3}}{1 + x'^{2} y^{2}} \frac{鈭� F}{鈭� x} = 0$

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

$鈭� y \sqrt{y^{2} + y'^{2}} d x = 鈭� y \sqrt{y^{2} + y'^{2}} x' d y = 鈭� y \sqrt{y^{2} + x'^{- 2}} x' d y = 鈭� \sqrt{x'^{2} y^{2} + 1} d y$

Let $F (x, y, y') = y \sqrt{x'^{2} y^{2} + 1}$ .

By the definition of the Euler equation, $\frac{d}{d y} \frac{鈭� F}{鈭� x'} - \frac{蟺贵}{鈭� x} = 0$ .

Differentiate F with respect to x' and x.

$\frac{鈭� F}{鈭� x'} = \frac{x' y^{3}}{\sqrt{1 + x'^{2} y^{2}}} \frac{鈭� F}{鈭� x} = 0$

The Euler becomes as shown below.

$\frac{d}{d x} [\frac{x' y^{3}}{\sqrt{1 + x'^{2} y^{2}}}] = 0 \frac{x' y^{3}}{\sqrt{1 + x'^{2} y^{2}}} = C$

The differential equation is $x' = \frac{C}{y \sqrt{y^{4} - C^{2}}}$

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

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Find the Lagrangian and the Lagrange equation for the pendulum shown. The vertical circle is fixed. The string winds up or unwinds as the massswings back and forth. Assume that the unwound part of the string at any time is in a straight-line tangent to the circle. Letbe the length of the unwound string when the pendulum hangs straight down.

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