Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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).

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

Step-by-Step Solution

Answer

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

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;

$$\begin{gathered} y\text{'}=v \\ y\text{'}\text{'}=-y^3 \\ v\text{'}=-y^3 \end{gathered}$$ 

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;

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

Step 2: Sketch

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