Timeline for Proper way to produce StreamPlot
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 29, 2020 at 18:26 | answer | added | Michael E2 | timeline score: 2 | |
| Mar 22, 2016 at 1:06 | comment | added | Moo | You can set $y = v x$ and solve $y' = \dfrac{x}{-x-2y}$ and then you can see why the PP is correct. | |
| Mar 21, 2016 at 18:12 | comment | added | RuD_wow | @Moo But, what if we will calculate the eigenvalues of system $\dfrac{dx}{dt}, \dfrac{dy}{dt} $ and, considering their Real values, make a conclusion that DE' s phase trajectories are logarithmic clockwise spiral? | |
| Mar 20, 2016 at 14:22 | comment | added | Moo | @RuD_wow: The denominator of that ODE cannot be $y = -\dfrac{x}{2}$, else the ODE is undefined. This is a line as shown in the phase portrait, so the phase portrait you show above is is correct. | |
| Mar 20, 2016 at 13:44 | comment | added | RuD_wow | @Rahul This one: $ y' = \dfrac{x}{-x-2y} $ | |
| Mar 19, 2016 at 1:34 | comment | added | user484 | What is the ODE you're trying to plot the trajectories of? | |
| Mar 19, 2016 at 0:43 | answer | added | Basheer Algohi | timeline score: 4 | |
| Mar 19, 2016 at 0:38 | comment | added | MathX | But are you sure this is not right? y=-x/2 is where the direction of tangent changes. Just plug in x=-1, y=1 above this line and x=-1 y=0 below this line, and you get the slope of +1 for the former case and -1 for the latter case. | |
| Mar 18, 2016 at 23:58 | history | asked | RuD_wow | CC BY-SA 3.0 |