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    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?
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    (Original post by specimenz)
    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?
    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} as 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' or y' has a constant in its expression, but this is the analysis I'd expect
 
 
 
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