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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).$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}{\mathbf{+}}{\mathbit{y}}{\mathbf{=}}{\mathbf{0}}$

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

See the step by step solution

## Step 1: Find the critical point

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

Put $v=y\text{'} a\mathrm{nd} v\text{'}=y\text{'}\text{'}$

Then the given system can be written as:

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

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

This is the required result.