# find equilibrium points for an infected population IVP

Let $x$ be the proportion of a population with a disease, let $y$ be the healthy ones

the disease rate spreads at $$\frac{dy}{dt} = ay(1-y),\quad y(0)=y_0$$

• $(a)$ find the equilibrium points for the differential equation and determine whether each is asymptotically stable, semistable, or unstable

I know the equilibrium solutions are $y=0$ and $y=1$. How do I determine their stability?

• $(b)$ solve the IVP and verify that part $(a)$ is correct

my solution is $$y = \frac{\frac{y_0}{y_0-1}e^{rt}}{\frac{y_0}{y_0-1}e^{rt}-1}$$

is this correct?

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Equilibrium points are solutions $y(t)$ for which $y'(t) = 0$ for all times. You can usually read the equilibria of an ODE of the form $x' = f(x)$ by determining the roots of $f$. Such roots are precisely the initial conditions that stay where they are under the passage of time. –  A Blumenthal Jan 31 '13 at 20:36

If you have an autonomous differential equation of the form $$\frac{dy}{dt}=f(y),$$ and you need to check the stability of equilibrium point $\hat{y}$, then you can use the following criterion:
• If $f'(\hat{y})>0$ then the equilibrium is undtable
• If $f'(\hat{y})<0$ then the equilibrium is asymptotically stable
• If $f'(\hat{y})=0$ then additional analysis is required.