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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 19–24, convert the given second-order equation into the first-order system by setting v=y’. Then find all the critical points in the yv-plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Figure 5.12).$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{+}}{\mathbit{y}}{\mathbf{+}}{{\mathbit{y}}}^{{\mathbf{5}}}{\mathbf{=}}{\mathbf{0}}$

The point is an unstable saddle point (0, 0).

See the step by step solution

## Step 1: Find a critical point

Here the equation is $\frac{{d}^{2}y}{d{t}^{2}}+y+{y}^{5}=0$.

Put $v=y\text{'} \mathrm{and} v\text{'}=y\text{'}\text{'}$.

Then the system is;

$y\text{'}=v\phantom{\rule{0ex}{0ex}}y\text{'}\text{'}=-y\phantom{\rule{0ex}{0ex}}v\text{'}=-y$

For critical points equate the system equal to zero.

$v=0\phantom{\rule{0ex}{0ex}}-y=0\phantom{\rule{0ex}{0ex}}y=0$

So, the critical point is (0, 0).

The phase plane equation is:

$\frac{dv}{dy}=\frac{-y}{v}\phantom{\rule{0ex}{0ex}}\int vdv=\int -ydy\phantom{\rule{0ex}{0ex}}{v}^{2}+{y}^{2}=c$

## Step 2: Sketch Therefore, the point is an unstable saddle point (0,0).

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