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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 19–24, convert the given second-order equation into a 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).$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{+}}{{\mathbit{y}}}^{{\mathbf{3}}}{\mathbf{=}}{\mathbf{0}}$.

The point is the center point (0, 0).

See the step by step solution

## Step 1: Find the critical point

Here the equation is $\frac{{d}^{2}y}{d{t}^{2}}+{y}^{3}=0$.

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

Then the system is;

$y\text{'}=v\phantom{\rule{0ex}{0ex}}y\text{'}\text{'}=-{y}^{3}\phantom{\rule{0ex}{0ex}}v\text{'}=-{y}^{3}$

For critical points equate the system equal to zero.

$\begin{array}{rcl}v& =& 0\\ -{y}^{3}& =& 0\\ y& =& 0\end{array}$

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

The phase plane equation is;

$\begin{array}{rcl}\frac{dv}{dy}& =& \frac{-y}{v}\\ \int vdv& =& \int -{y}^{3}dy\\ \frac{{v}^{2}}{2}& =& -\frac{{y}^{4}}{4}+c\\ 4{v}^{2}+{y}^{2}& =& c\end{array}$

## Step 2: Sketch

Therefore, the point is the center point (0, 0).