Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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).

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

Step-by-Step Solution

Answer

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

Step 1: Find a critical point

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

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

 

Then the system is;

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

For critical points equate the system equal to zero.

$$\begin{gathered} v=0 \\ -y=0 \\ y=0 \end{gathered}$$

 

So, the critical point is (0, 0). 

The phase plane equation is:

$$\begin{gathered} \frac{dv}{dy}=\frac{-y}{v} \\ \int vdv=\int -ydy \\ {v}^2+y^2=c \end{gathered}$$

 

Step 2: Sketch

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