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I want to find the equilibrium points of this differential equation:


i have found that $y=\frac{(a-b)}{a }$makes $y'(t)=0$ but i don't know if $y'(t)=0$ is also an equilibrium solution

thanks for the help!

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yes your approach is true . you can calculate$ y''$ to find out type of equilibrium points (sink or source) if $y''(x_e)<0 then $x_e$is absorb point and $y''(x_e).0 then $x_e$is source point – Maisam Hedyelloo Feb 3 '13 at 16:20
up vote 1 down vote accepted

Any value of $y$ that makes $y'=0$ is an equilibrium point. If $ay(1-y)-by=0$ then either $y=0$ or $a(1-y)-b=0$. The latter implies $y=(a-b)/a$. So $0$ and $(a-b)/a$ are both equilibrium points, and there are no others.

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An equilibrium solution is a constant solution to a differential equation. If you draw a slope field, the equilibrium solution is a horizontal line . So if you'd like to find the equilibrium solution for an OE, you have to put the OE equal to zero and solving for the variable value. Exactly what @Micheal did in details.

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Nice overview of the problem and approach to solving it +1 – amWhy Feb 4 '13 at 1:02

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