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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

# In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).$\frac{\mathbf{dx}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{\left(}}{\mathbit{y}}{\mathbf{-}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{\left(}}{\mathbit{y}}{\mathbf{-}}{\mathbf{1}}{\mathbf{\right)}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{-}}{\mathbit{y}}{\mathbf{\right)}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{-}}{\mathbf{1}}{\mathbf{\right)}}$

The solution is $\mathbf{\left(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{\right)}}^{\mathbf{2}}{\mathbf{+}}\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{1}{\mathbf{\right)}}^{\mathbf{2}}{\mathbf{=}}{\mathbit{c}}$.

See the step by step solution

## Step 1: Find phase plane equation

Here the system is:

$\frac{dx}{dt}=\left(y-x\right)\left(y-1\right)\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=\left(x-y\right)\left(x-1\right)$

And the phase plane equation is:

$\frac{dy}{dx}=\frac{\left(x-y\right)\left(x-1\right)}{\left(y-x\right)\left(y-1\right)}\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=\frac{1-x}{y-1}$

## Step 2: Solve the equation

Here the equation is $\frac{dy}{dx}=\frac{1-x}{y-1}$.

Solving by variable separating. Then,

$\int \left(y-1\right)dy=\int -\left(x-1\right)dx\phantom{\rule{0ex}{0ex}}{\left(y-1\right)}^{2}+{\left(x-1\right)}^{2}=c$

Since the solutions are centered at (1,1). And line y = x.

## Step 3: Sketch some trajectories.

Therefore, the solution is $\mathbf{\left(}\mathbf{y}\mathbf{-}\mathbf{1}{\mathbf{\right)}}^{\mathbf{2}}{\mathbf{+}}\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{1}{\mathbf{\right)}}^{\mathbf{2}}{\mathbf{=}}{\mathbit{c}}$.