How to Determine the Direction of Curvature in a Phase Portrait?

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Homework Statement


Get the actual solution for the differential equation in matrix form? It is uploaded


Homework Equations





The Attempt at a Solution


My question is why is it drawn in the upload as the top picture on the upload shows, why not as the second picture shows how did they figure out which way the curvature goes? I realize the positive eigenvalues indicate that the two eigenvectors marked in dark black will go away from the origin and thus so will the curveing line. I can also see that (ada1) vector is bigger and so as T values are positive and large it will dominate and go parallel to (ada2) and when negative and large T are input it will go parallel to (ada1).
 

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This is where I got the problem from it may be clearer here: http://www.math.lamar.edu/faculty/dawkins/archive/Spring2011/3301/11SpringHwk10.pdf . and the next pdf is for the solutions: http://www.math.lamar.edu/faculty/dawkins/archive/Spring2011/3301/11SpringHwk10_Soln.pdf . If you can't access this please go to lamar.edu click academics click mathematics then faculty and staff then paul dawkins then archives then 2011 and spring semester go to homework 10.
 
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What you have done is very good. Note that the eigenvector <7, 4>, which gives to the line y= (4/7)x, corresponds to eigenvalue +4 while the eigenvector <1, 1>, which gives the line y= x, corresponds to eigenvalue +1. Since both eigenvalues are positive all phase lines move outward from the origin but tend to go more along y= (4/7)x than along y= x because that has the larger eigenvalue. In your first picture, the lines do tend first along y= (4/7)x, some then turning back toward y= x. In your second picture, where you ask "why not this" is the other way.

So there is your answer. The first picture is correct because eigenvalue 4 is larger than eigenvalue 1.
 
Ok I think I have it now to must go with the eigenvalue that is larger and I can kinda see why that makes sense because the lines can't cross each other so for it to go with the larger of the two there is only one way it can go without crossing but what if I get a problem where I can do it both ways such that it follows the larger eigenvalue?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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