the function is dy/dy=y^2+3y+a and after using the quadratic formula, I get an equilibrium point of a=9/4. Now when I set up the phase diagrams, I have a<9/4, a=9/4, and a>9/4. I know that in the order I just described they will have 2 roots, 1 root, and no roots. The problem I am having is drawing the direction of the arrows. Am I picking points and plugging them into the quadratic formula? If anyone can clarify this step, it would be much appreciated.
drawing bifurcation/phase diagram for dy/dy=y^2+3y+a
- Thread starter lbros55
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Dr.Peterson
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I'm not entirely sure what specific sort of diagrams you are drawing; these terms are used in somewhat different settings. This page may be what you are talking about. There, in a picture about halfway down, they have arrows (in red) pointing up where the derivative (right side of your equation) is positive, and (in blue) pointing down where it is negative. I suppose that could be determined just by plugging in coordinates of a point inside the curve and seeing whether it is positive or negative; but I would use the fact that y^2+3y+a is concave up, so that if it is negative anywhere, it will be in the middle (between the two zeros). So this will be the reverse of the example there (where p(1-p) is concave down, making it positive only in the middle). Understanding the overall behavior is always better than just blindly trying something.
But checking a point is still a good idea, to make sure you are right.