I have a differential equation $$ 0.25x'+0.0001x''+0.0001x=\sin(2000\pi t)$$ with initial conditions $x(0)=0, x'(0)=0$.

WolframAlpha plots the solution to this as

WolframAlpha solution

but if I plot it on this website using the following system of equations:


I get this phase portrait which I don't see how relates to my solution.

phase portrait

  • $\begingroup$ In the first case, the axes are (t,x), in the second case, the axes are (x,y)... $\endgroup$
    – Jean Marie
    Apr 24 at 12:13
  • 1
    $\begingroup$ "This page plots a system of differential equations of the form dx/dt = f(x,y), dy/dt = g(x,y)." So you can't have $t$ on the right-hand side! (The concept of phase portrait makes no sense for non-autonomous systems.) $\endgroup$ Apr 24 at 12:47
  • $\begingroup$ @HansLundmark Oh well that's not good... I guess I'm not quite sure what I thought a phase portrait was then. Thanks for clearing this up for me a bit. $\endgroup$
    – Matt
    Apr 24 at 13:11

Considering the ODE

$$ 0.25x'+0.0001x''+0.0001x=\sin(2000\pi t) $$

or normalizing

$$ x''+2500x'+x=10000\sin(2000\pi t) $$

we can construct the first order equivalent system

$$ \cases{ \dot x_1 = x_2\\ \dot x_2 = -2500x_2-x_1 -10000\sin(2000\pi t) } $$

Considering the corresponding homogeneous/autonomous system

$$ \cases{ \dot x_1 = x_2\\ \dot x_2 = -2500x_2-x_1 } $$

we can draw for the autonomous part, a phase portrait as follows:

enter image description here

From this phase portrait we can conclude that the autonomous system is stable to initial conditions or in other words, given $\{x_1(0), x_2(0)\}\in \mathbb{R}^2$ the system evolves to the origin.


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