draw a bifurcation diagram
x' = x^3-x+lambda where lambda is a parameter.

Question

draw a bifurcation diagram
x' = x^3-x+lambda where lambda is a parameter.

usukidoll · Answer

I need the phase lines to draw a bifurcation diagram right?!

usukidoll · Answer

I swear I want to destroy this thing. -_- so what the heck do I need if I don't have to find equilibria, stability, and phase lines then -_-

usukidoll · Answer

@dan815 destroy this with me please ^_^

usukidoll · Answer

@perl any ideas?

usukidoll · Answer

let's destroy this cr**. XD Seriously, I showed this to my prof and he was like lol why do you need phase lines? r u effing kidding me? There was a tutorial that required the phase lines to draw the diagram omg

dan815 · Answer

no u dont

dan815 · Answer

u need the if u are a human

dan815 · Answer

them

usukidoll · Answer

huh?!

dan815 · Answer

with a computer u can numerically get all the plots

usukidoll · Answer

I'm a human, not a vocaloid XD.

usukidoll · Answer

yeah but let's say I need to do it manually incase an exam question has this.

usukidoll · Answer

sigh ... If I go his route..... taking the derivative of that I will have 3x^2-1
then I will have 3x^2-1=x^3-x+lambda
which in turn becomes
3x^2-1-x^3+x=lambda
-x^3+3x^2+x-1=lambda

that doesn't tell much...mind you I was sick for a week, so I missed the entire lecture.

usukidoll · Answer

-x^2(x-3)+x-1 
doesn't work through factor by grouping either. ugh

usukidoll · Answer

this sucks :/.
difference equations are better.

dan815 · Answer

if u wanna do this manually then

dan815 · Answer

partial fractions

usukidoll · Answer

daphuc?

dan815 · Answer

|dw:1423556607097:dw|