Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

The only critical point is (0,0).

Step 1: Find critical points

Consider the system as;

 $$\begin{gathered} \frac{dx}{dt}=x^2-2xy \\ \frac{dy}{dt}=3xy-y^2 \end{gathered}$$ 

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

So, 

$$\begin{gathered} x^2-2xy=0 \\ 3xy-y^2=0 \end{gathered}$$

Step 2: Solve for x and y

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

 $\begin{array}{rcl}x(x-2y)& =& 0\\ x& =& 0 or 2y\\ y(3x-y)& =& 0\\ y& =& 0 or 3x\end{array}$ 

Thus, the only critical points are (0,0). 

This is the required result.