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Differential Equation Phase Plane Portrait.

So given a non homogenous system of ode:

x'=-x+6y -2
y'=-x+4y

sketch the phase portrait.

So I first calculated the eigen values/vectors of the matrix
|-1 6|
|-1 4|
and find for lambda=2, eigen vector = (2,1)
lambda=1, eigen vector = (3,1)

and particular solution = (4,1)

How do I sketch the phase portrait?
phase portrait.png

Reply 1

RDKGames

Study Forum Helper

Original post by specimenz

Well your particular solution is your critical point, first of all.

Then note that you have

\displaystyle \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + C_1 \begin{pmatrix} 3 \\ 1 \end{pmatrix} e^{t} + C_2 \begin{pmatrix} 2 \\ 1 \end{pmatrix} e^{2t}

for some constants

C_1,C_2

Now, as

t \rightarrow -\infty

we have

(x,y) \rightarrow (4,1)

which means that every arrow will be coming from the critical point (so it's unstable), and furthermore,

e^t >> e^{2t}

t \rightarrow - \infty

therefore every vector will come out parallel to the eigenvector

(3,1)

(your

\lambda = 1

).

Now consider

t \rightarrow \infty

, which means that

(x,y) \rightarrow (\infty, \infty)

and more specifically, they will both tend to infinity along the vector

(2,1)

because at +ve infinity

e^{2t} >> e^{t}

so that term takes control.

The rest is just sketching accordingly to this.

Small disclaimer: As I'm learning this topic at the moment, I haven't come across the case yet where

x'

y'

has a constant in its expression, but this is the analysis I'd expect

(edited 6 years ago)

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