# What is the meaning of equilibrium solution?

What are the equilibrium solutions for the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}t} = 0.2\left(y-3\right)\left(y+2\right)$

My Question: What does equilibrium solution mean in this context of this problem?

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It's the solution to which other solution tend to approach or in opposite moves away when $t\rightarrow\infty$. It's equilibrium because at those points solution doesn't depend on time (or any other variable you're integrating over). Those solutions that "attracts" other ones that started near them are called stable. Those that "push away" solutions that started near them are called unstable. And finally, in some cases there are solutions that either attracts or pushes other solutions depending on which side from them other solutions started are called semi-stable.

For example there are several such lines on the picture below.

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They're the points in phase space at which the solution does not change. If you imagine your solution $y$ as a point-mass which moves in time, then the equilibrium solutions are the points in phase space (where your particle moves) at which the particle has zero velocity. Hence, if your solution has an initial condition which places it at such a point, then the solution remains constant for all time (because there is no additional inhomogeneity [e.g. forcing function] in which to "move" or accelerate the particle). Note also, that by uniqueness (assuming the usual regularity conditions), the constant solution at such an equilibrium point is the only solution, so if you don't start at an equilibrium solution, your particle will never arrive at a point of being constant for all later times (possible behaviors include oscillations, going to infinity, blowing up, asymptotically approaching equilibrium, etc.).

In this case, $y=3$ and $y=-2$ are the equilibrium points of this system. So solutions corresponding to $y(0)=3$ and $y(0)=-2$ are actually equal to $3$ and $-2$, respectively, for all time $t$ (note that because the system is autonomous (no inhomogeneity), it is unimportant that the initial time is $t=0$ -- anyway, the concept of equilibrium point loses some importance/meaning when one considers non-autonomous systems).

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The equilibrium solutions are values of $y$ for which the differential equation says $\dfrac{dy}{dt} = 0$. Therefore there are constant solutions at those values of $y$.

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