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Q13E

Expert-verifiedFound in: Page 271

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).**

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

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

Here the system is:

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

And the phase plane equation is:

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

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.

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

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