# Critical points of differential equation

1. Jul 10, 2012

### samee

1. The problem statement, all variables and given/known data

Determine the location and type of all critical points of the given equations and sketch the phase portrait

y"+cosy=0

3. The attempt at a solution

I've done some like this before but they were all systems of equations. I'm actually not sure how to do the simpler version >_< I think this is asking for maxima and minima? is that right?

2. Jul 10, 2012

### HallsofIvy

Staff Emeritus
It's NOT simpler. This is only one equation but is second order. What you want is two first order equations. Define x by x= y'. Then y''= x' so y''+ y= 0 becomes x'+ y= 0 or x'= -y.

You now have the two equations x'= -y and y'= x.

3. Jul 10, 2012

### samee

Okay, so for my problem I have
x=y'
x'=y"
so I substitute and
x'+cosy=0
x'=-cosy

so my system of equations is;

y'=x
x'=-cosy

right? Then I just solve like it's a system of equations and look for the singularities as the critical points?

4. Jul 10, 2012

### samee

Ah! No wait, there's more! I know what I'm doing now, silly me. I set x' and y' to zero and solve for the points.

x'=0, ∴x=0
y'=0, -cosy=0, ∴y=(1/2)(2n+1)pi

So I have infinite points along the y-axis. I used

http://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html [Broken]

to graph the phase portrait and found that every other critical point is a saddle point, then a center point.

SO! how do I say that one is a saddle, then the next is a center?

Last edited by a moderator: May 6, 2017