# Thread: Phase Portrait using Isoclines?

1. ## Phase Portrait using Isoclines?

I want to sketch the phase portrait (solution curves) of this DE

$y'=y-x^2$

Could I use isoclines? i.e. something like this

$0=y-x^2$
$=>y=x^2$

So i would sketch a parabola with slope 0 hairs on it? Then do the same thing for slope 1.

$1=y-x^2$
$=>y=\frac{1}{x^2}$

So i would sketch 1/x^2 with slope 1 hairs on the curve?

2. You can just plot small vectors at a rectangular grid of numbers in the x-y plane say from -5 to 5, the slope of the vectors of course is the value $y-x^2$ at each point so for example at the point (1,1), I'd draw a small line with a slope of 0. At the point (0,2), the line would have a slope of 2 and so forth. Once you understand the meaning of that, then try StreamPlot in Mathematica which plots the vector field of {x'(t), y'(t)} so if I let x=t then I'd need to plot the field {1,y-t^2} like so:

StreamPlot[{1, y - t^2}, {t, -5, 5}, {y, -5, 5}] and that gives me the phase portrait below which I've also drawn a solution (in red) to the DE for y(0)=1.

Also check out "Differential Equations" by Blanchard, Devaney, and Hall. It's the best reference for these types of problems.