Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

There is another exercise where I am given a non-linear system of ODEs and a question that confuses me:

Consider $$x'=x(y+1)\\ \text{ and }\\ y'=xy+2.$$

"Determine stationary points of the system, and linearize the system at the stationary points, determine the geometric type of the linearizations and the stability."

The only stationary point is $(x=2,y=-1)$. For the stability, I wrote $$f(x)=x(y+1),\\ g(y)=xy+2,$$ then differentiated: $$f'(x)=y+1,\\ g'(y)=x.$$ At $(-1,2)$, these values are $0$ and $2$. For $f$, I wrote that we cannot use linear terms in the expansion and hence would have to consider $f''$ and so on. For $g$, since it's greater than zero, the point is not linearly stable in $y$-direction.

But what is meant by "determine the geometric type of the linearizations"? What are types of linearizations? What am I missing here?

all the best,


share|cite|improve this question
up vote 1 down vote accepted

For the stability you should evaluate the Jacobi matrix (see my comment here). The geometric type of linearization is determined by the eigenvalues of this matrix. You can start learning these matters, e.g., here.

share|cite|improve this answer

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.