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Q22E

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 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).**

**$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{+}}{{\mathit{y}}}^{{\mathbf{3}}}{\mathbf{=}}{\mathbf{0}}$**.

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

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

Put $v=y\text{'}\u200a\u200a\mathrm{and}\u200a\u200av\text{'}=y\text{'}\text{'}$.

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;

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

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

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