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Find the critical value or values of alpha where the qualitative nature of the phase portrait for x'=(2, alpha, -5, -2)x changes. (this is 2x2 matrix, 2 and alpha on the left, -5 and -2 on the right.)


Quick comment to Hassan's solution:
You have either two purely imaginary eigenvalues or two purely real eigenvalues or degenerate zero eigenvalue. If two eigenvalues are purely imaginary, your solution will have exponents with imaginary power, which reduce to sine and cosine functions. Your solution is bound and any fixed points your system has are stable.
If two roots are real, one is positive and one is negative. Your solution will have e to the positive and e to the negative power, that is exponentially decreasing and exponentially increasing solution. In this case the solution is unbound and the fixed points your system might have are unstable.
As a quick addition to Kirill's comment to Hassan's solution: we must not forget the critical case, where r=0 is the only eigenvalue. In this case your two independent solutions are a constant, corresponding to a stable fixed point in the phase diagram, and a linear function, corresponding to an unbounded phase line in the direction of the eigenvector.

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Hassan H. | Math Tutor (All Levels)Math Tutor (All Levels)
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Hello Sun,
Just for your reference (and since I don't have too much time right now), first find the eigenvalues of the matrix.  Do this by solving
(2 - r)(-2 - r) + 5α = 0,
the solutions will be your eigenvalues (r = ±√(4-5α) for your reference).  To find the critical values, simply locate where the solutions will change from complex to real---in this case, α = 4/5.  If you need help on how to interpret the qualitative nature of he phase portrait, I'm sure some of the other hyperactive tutors will come by and complete the picture for you, so to speak, maybe Kirill or Andre or Robert (just a friendly poke since I see their names often on the boards).
Hassan H.