- #1

- 22

- 0

I'm trying to determine how

y' = sin(y)

is different from

y' = 2 + sin(y)

I plotted them both in Converge and I don't understand how adding a two rotated and skewed the graph.

Can you explain to me however possible what is going on here?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter epheterson
- Start date

- #1

- 22

- 0

I'm trying to determine how

y' = sin(y)

is different from

y' = 2 + sin(y)

I plotted them both in Converge and I don't understand how adding a two rotated and skewed the graph.

Can you explain to me however possible what is going on here?

- #2

- 22

- 0

Or check out the attachment, I watched it while adding increments of .5 until I got up to y'=sin(y) + 2

It seems like every time you add a greater number, you increase the slope a little more. And past the point of y=sin(y) + 1, there are no more equilibrium solutions, it becomes monotonous.

Help?

I attached a word document with all the graphs from Converge

It seems like every time you add a greater number, you increase the slope a little more. And past the point of y=sin(y) + 1, there are no more equilibrium solutions, it becomes monotonous.

Help?

I attached a word document with all the graphs from Converge

- #3

- 287

- 0

It will rotate it because every slope becomes steeper, correct? Slopes that were 1 are now 3. That's a significant rotation.

It's looking translated is probably a coincidence. What really happened was an increasing of the slopes... and not a uniform one, at that.

So it's not really rotating or translating, although the periodicity of the field and the and the odd field transformation is making it look like that.

- #4

- 22

- 0

I understand completely now, thanks

Share:

- Replies
- 3

- Views
- 5K

- Replies
- 1

- Views
- 2K