Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3–6, find the critical point set for the given system.

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{+}\mathbf{2}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{y}\mathbf{-}\mathbf{2}\mathbf{)}$

Step-by-Step Solution

Answer

The only critical points are (1, 1).

Step 1: Find critical points

Consider the system as;

 $$\begin{gathered} \frac{dx}{dt}=y^2-3y+2 \\ \frac{dy}{dt}=(x-1)(y-2) \end{gathered}$$ 

For finding the critical points put the system equal to 0. 

So, 

 $$\begin{gathered} y^2-3y+2=0 \\ (x-1)(y-2)=0 \end{gathered}$$

Step 2: Solve for x and y

Put the value of y in another equation and solve, then;

 $\begin{array}{rcl}(y-2)(y-1)& =& 0\\ y& =& 2 or 1\\ (x-1)(y-2)& =& 0\\ x& =& 1,y=2\end{array}$ 

Thus, the only critical points are (1,1). 

This is the required result.