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Expert-verified 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 # In Problems 3–6, find the critical point set for the given system.$\frac{\mathbf{dx}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{,}}\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{=}}{{\mathbf{x}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{y}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{1}}$

The critical points are $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

See the step by step solution

## Step 1: Find critical points

Consider the system as:

$\frac{dx}{dt}=x-y\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}={x}^{2}+{y}^{2}-1$

For finding the critical points put the system equal to 0.

So,

$\begin{array}{rcl}x-y& =& 0\\ x& =& y\\ {x}^{2}+{y}^{2}-1& =& 0\\ {x}^{2}+{y}^{2}& =& 1\end{array}$

## Step 2: Solve for x and y

Put the value of y in another equation and solve, then;

$\begin{array}{rcl}{x}^{2}+{x}^{2}& =& 1\\ 2{x}^{2}& =& 1\\ x& =& ±\frac{1}{\sqrt{2}}\end{array}$

Thus, the critical points are $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right){\mathbf{,}}\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$.

This is the required result.

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