Q23E - 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).

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{-}\mathit{y}\mathbf{(}\mathbf{t}{\mathbf{)}}^{\mathbf{4}}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The critical points are (0,0),(1,0).

Step 1: Find the critical point

Here the equation is $y\text{'}\text{'}\left(t\right)+y\left(t\right)-y{\left(t\right)}^4=0$.

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

 

Then the system is;

 $$\begin{gathered} y\text{'}=v \\ y\text{'}\text{'}=-y+y^4 \\ v\text{'}=-y+y^4 \end{gathered}$$

 

For critical points equate the system equal to zero.

 $\begin{array}{rcl}v& =& 0\\ -y+y^4& =& 0\\ y\left(-1+y^3\right)& =& 0\\ y& =& 0 or \left(-1+y^3\right)=0\end{array}$ 

If $y\ne 0 \text{then}$

$\begin{array}{rcl}-1+y^3& =& 0\\ y& =& 1\end{array}$

So, the critical point is (0, 0) and (1, 0).

Step 2: Sketch the directional field.

Therefore, the critical points are (0, 0), (1, 0).