Finding Critical Points of y1'=y2 and y2'=-kcos(y1)

[helpinghand]

Hey Guys,

I need help finding the critical points of this system:

y₁^'=y₂ ..... 1
y₂^'=-kcos(y₁) ..... 2

Critical Points (y₁, y₂)

For eqn 1, would the CP be (0,0)?

For eqn 2, how would I find out what the CP is?

Any help would be awesome.

Thanks

 

[HallsofIvy]

You don't find "critical points" of individual equations in a system. A critical point for a system of first order differential equations is a point so that all of the derivatives are 0.

To be a critical point for this system you must have y2= 0 and -kcos(y1)= 0. cos(y1)= 0 if and only if y1 is an odd multiple of \pi/2.