Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.
share|cite|improve this answer

Stability is determined by the set of initial conditions that will lead asymptotically to the given equilibrium point. For this diff eq, ALL initial conditions less than zero tend to zero, ALL between zero and one tend to one, and all greater than one tend to one. You can tell this from a slope field using simple observation and the existence and uniqueness theorem.

share|cite|improve this answer
"ALL initial conditions less than zero tend to" -oo. – Did Jul 1 '14 at 19:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.