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Q21E

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 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).**

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{+}}{\mathit{y}}{\mathbf{+}}{{\mathit{y}}}^{{\mathbf{5}}}{\mathbf{=}}{\mathbf{0}}$

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

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

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

Then the system is;

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

**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:

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

Therefore, the point is an unstable saddle point (0,0).

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