# How does my phase portrait fit with my differential equation?

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

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.

• In the first case, the axes are (t,x), in the second case, the axes are (x,y)... Apr 24 at 12:13
• "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.) Apr 24 at 12:47
• @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.
– 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:

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.