I solved an initial value problem and got the following:
$x(t)=e^{2t}$
$y(t)=e^{-3t}$
So, the equilibrium point $(0,0)$ is of saddle nature.
Also $x(t)\rightarrow+\infty$ and $y(t)\rightarrow 0$ as $t\rightarrow\infty$
So the phase portrait should be like this:
Is it correct? Thanks for any response.
EDIT
The original problem statement is:
Determine the nature of equilibrium point (0,0) of the system $\dot x=x+y, \dot y=4x-2y$ subject to the initial condition $(x(0),y(0))=(2,-3)$. Also sketch the phase portrait.
Best Answer
We have to determine the nature of equilibrium point $(0,0)$ of the system $$\begin{align} x' &= x+y\\ y' &=4x-2y \end{align}$$
subject to the initial condition $(x(0),y(0))=(2,-3)$ and sketch the phase portrait.
We can find the critical points as $(0,0)$ by simultaneously solving
$$x+y = 0\\ 4x-2y = 0$$
If we find the eigenvalues of
$$\begin{pmatrix} 1 & 1 \\ 4 & -2 \end{pmatrix}$$
We get $$\lambda_1 = -3, \lambda_2 = 2$$
This already tells us that we have a saddle-point, which is unstable.
When we solve this system, we get
$$\begin{align} x(t) &= e^{-3 t}+e^{2 t} \\ y(t) &= -4 e^{-3 t} + e^{2 t} \end{align}$$
If we draw a phase portrait and then superimpose the initial condition solution (red item), we get
Update
If you mean, for
$$\begin{pmatrix} -3 & 0 \\ 0 & 2 \end{pmatrix}$$
We get