Dy/dt = sin^2 y ebilibrium

[killersanta]

Homework Statement

I need to find the equilibrium solutions
dy/dt = sin^2 y

The Attempt at a Solution

I don't know what to do with the sin^2. If it was just sin, it would be easy, 0, pi, -pi, 2pi, -2pi...etc... But it's not, I have a Ti-84 plus calculator. So, i don't believe I can put it in my calculator to find out? what do I do?

 

[Tangent87]

If you're looking for the points where (siny)^2 is zero, surely these are just the same points as siny=0?

 

[killersanta]

Thanks, I was kind of thinking that. There was just no way for me to put it in my calculator to check.

 

[killersanta]

After I find the equilibrium points I have to determine if they're a sink, source or a node. Since I can't put it in my calculator, how do I determine that? Are they the same as sin y or the opposite?

 

[fzero]

After I find the equilibrium points I have to determine if they're a sink, source or a node. Since I can't put it in my calculator, how do I determine that? Are they the same as sin y or the opposite?

Instead of worrying about what your calculator can do, you might want to review your notes and text for the definitions of sink, source and node. If you're having trouble understanding those notions, post back here so that someone can help explain them.

 

[killersanta]

I know what a sink, source and node are. Sink, both sides are going to it. Source, going away, node is both. Don't you have to put it in your calculator to see what they are doing between equilibrium's?

 

[Char. Limit]

sin^2(x) is pretty easy to imagine. It's a lot like sin(x), but it's always positive. (Remember that it's also 1-cos^2(x), which can easily be shown to be always positive).

 

[killersanta]

Thanks, that make senses. So basically the equilibrium points are the same as sin y. -2pi, -pi, 0, pi, 2pi...etc but they are all nodes due to being always positive because sin^2 y = 1-cos^2 y.