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Found in: Page 271

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Find all the critical points of the system$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{1}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbit{x}}{\mathbit{y}}$and the solution curves for the related phase plane differential equation. Thereby proving that two trajectories lie on semicircles. What are the endpoints of the semicircles?

The result is

$for y>0,\left|x\right|>1,y={e}^{c}\sqrt{{x}^{2}-1}\phantom{\rule{0ex}{0ex}}for y>0,\left|x\right|<1,y={e}^{c}\sqrt{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}for y<0,\left|x\right|>1,y=-{e}^{c}\sqrt{{x}^{2}-1}\phantom{\rule{0ex}{0ex}}for y<0,\left|x\right|<1,y=-{e}^{c}\sqrt{1-{x}^{2}}$

And the end points are (-1,0) (1,0).

See the step by step solution

## Step 1: Find critical points

For a critical point put the system equal to 0.

${x}^{2}-1=0\phantom{\rule{0ex}{0ex}}xy=0\phantom{\rule{0ex}{0ex}}{x}^{2}=1\phantom{\rule{0ex}{0ex}}x=±1\phantom{\rule{0ex}{0ex}}y=0$

## Step 2: Find the value of y

Now,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{xy}{{x}^{2}-1}\\ \int \frac{1}{y}dy& =& \int \frac{{x}^{2}}{{x}^{2}-1}\\ \mathrm{ln}\left|y\right|& =& \frac{1}{2}\mathrm{ln}\left|t\right|+c \left(\text{bysubtitution}\right)\\ \mathrm{ln}\left|y\right|& =& \mathrm{ln}{\left|t\right|}^{\frac{1}{2}}+c\\ y& =& {e}^{c}{\left|t\right|}^{\frac{1}{2}}\\ y& =& {e}^{c}\sqrt{{x}^{2}-1}\end{array}$

## Step 3: Prove results that endpoints are semicircles.

Now there are two cases when y > 0 and y < 0 for both cases ${x}^{2}-1<0 \mathrm{and} {x}^{2}-1>0$ then $\left|x\right|<1 and \left|x\right|>1$

$fory>0,\left|x\right|>1,y={e}^{c}\sqrt{{x}^{2}-1}\phantom{\rule{0ex}{0ex}}fory>0,\left|x\right|<1,y={e}^{c}\sqrt{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}fory<0,\left|x\right|>1,y=-{e}^{c}\sqrt{{x}^{2}-1}\phantom{\rule{0ex}{0ex}}fory<0,\left|x\right|<1,y=-{e}^{c}\sqrt{1-{x}^{2}}$

The trajectories that possibly lie on semicircles. If we square the equation then

${y}^{2}={e}^{2c}\left(1-{x}^{2}\right)$

This will be the equation of circle only if ${{\mathbit{e}}}^{\mathbf{2}\mathbf{c}}{\mathbf{=}}{\mathbf{1}}$. Therefore, only two solutions lie on the semicircle, ${\mathbit{y}}{\mathbf{=}}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{\mathbf{,}}{\mathbit{y}}{\mathbf{=}}{\mathbf{-}}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$.

Therefore, it is a circle of radius 1 center at origin the endpoints are$\left(-1,0\right), \left(1,0\right)$.