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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:
diagram

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.

Esha
  • 686

1 Answers1

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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

enter image description here

Update

If you mean, for

$$\begin{pmatrix} -3 & 0 \\ 0 & 2 \end{pmatrix}$$

We get

enter image description here

Moo
  • 11,311
  • Actually I have drawnthe phase portrait in the tranformed plane where the matrix is in Jordan form. Can you draw the phase portrait in that plane? – Esha Nov 07 '22 at 17:03
  • And where will be the red line i.e. superimposing the initial condition in the second diagram? – Esha Nov 07 '22 at 18:41
  • Well, if you transformed the system, wouldn't you have to transform the ICs? – Moo Nov 07 '22 at 18:43
  • Yes, that's what I got in my answer i.e. I got (1,1) as the new ICs – Esha Nov 07 '22 at 18:44
  • Assuming $(1,1)$ is correct, see updated diagram. – Moo Nov 07 '22 at 18:47
  • Can you please explain how are we getting that red line? – Esha Nov 08 '22 at 02:04
  • I overlay the solution using a parametric plot over the phase portrait. You can also solve the for $x$ and $y$ numerically and overlay that onto the phase portrait. – Moo Nov 08 '22 at 02:58
  • Another approach is to put the IC on the phase portrait and then follow the direction field of the $x$ and $y$ derivatives. – Moo Nov 08 '22 at 03:04