Answer to Question #220713 in Differential Equations for Komal

Question

Find the nature and critical point of the system
dx/dt = (x)^2 +3xy+2(y)^2 and
dy/dt= 4-(x)^2

Answer 1

The only critical point is the solution of the simultaneous equations

$$(x)^2 +3xy+2(y)^2=0$$

$$4-(x)^2=0$$

$$x=-2: 4-6y+2y^2=0$$

$$y^2-3y+2=0$$

$$(y-1)(y-2)=0$$

$$Point(-2, 1), Point (-2, 2)$$

$$x=2: 4+6y+2y^2=0$$

$$y^2+3y+2=0$$

$$(y+1)(y+2)=0$$

$$Point(2, -2), Point (2, -1)$$

Critical points of the system

$$Point(-2, 1), Point (-2, 2),$$

$$Point(2, -2), Point (2, -1)$$

The Jacobian matrix is

$$J=\begin{pmatrix} \dfrac{\partial f}{\partial x} & \dfrac{\partial f}{\partial y} \\ \\ \dfrac{\partial g}{\partial x} & \dfrac{\partial g}{\partial y} \end{pmatrix}=\begin{pmatrix} 2x+3y & 3x+4y \\ -2x & 0 \end{pmatrix}$$

At the $Point(-2, 1)$  we get the matrix $\begin{pmatrix} -1 & -2 \\ 4& 0 \end{pmatrix}$ and so the two eigenvalues are 

$\dfrac{-1-i\sqrt{31}}{2}$ and $\dfrac{-1+i\sqrt{31}}{2}.$

As the eigenvalues are complex with negative real part, we get a spiral sink, which is an asymptotically stable equilibrium point.

At the $Point(-2, 2)$  we get the matrix $\begin{pmatrix} 2 & 2 \\ 4 & 0 \end{pmatrix}$ and so the two eigenvalues are $-2$ and $4.$

As the eigenvalues are real and opposite signs, we get a saddle, which is an unstable equilibrium point.

At the $Point(2, -2)$  we get the matrix $\begin{pmatrix} -2 & -2 \\ -4 & 0 \end{pmatrix}$ and so the two eigenvalues are $-4$ and $2.$

As the eigenvalues are real and opposite signs, we get a saddle, which is an unstable equilibrium point.

At the $Point(2, -1)$  we get the matrix $\begin{pmatrix} 1 & 2 \\ -4 & 0 \end{pmatrix}$ and so the two eigenvalues are 

$\dfrac{1-i\sqrt{31}}{2}$ and $\dfrac{1+i\sqrt{31}}{2}.$

As the eigenvalues are complex with positive real part, we get a spiral source, which is an unstable equilibrium point.