1. ## bifurcations

think i am on a role for not realising the obvious here...

i have a equationdy/dt= (1-y)(y^2-&)

&=lamda for the moment if i refer to it as that.

i need to find all birfucation points and sketchthe diagram. and also expand this by expanding f(y,&) usining taylors series.)

i have found critical points p1=(1,1)

and p2=(1/3,-1/3)

then found when fy <0 &<1

and fy>0 &>1
when y=1

when y= -1/3

i then dont know what i am doin please may you help

2. First solve for the equilibrium points:

$-y^3+y^2+y\lambda-\lambda=0$

I get $y=1$ and $y=\pm \sqrt{\lambda}$

So when $\lambda<0$ we have 1 (real) equilibrium point, when $\lambda=0$ we have two equilibrium points (a bifurcation occurs), when $\lambda>0$ another bifurcation occurs creating 3 equilibrium points.

How about I just show you the Mathematica code and the resulting plots for $\lambda=4$ and you try to figure it out:

Code:
myLambda = 4;
sol1 = NDSolve[{Derivative[1][y][x] == -y[x]^3 + y[x]^2 + y[x]*\[Lambda] - \[Lambda] /.
\[Lambda] -> myLambda, y[0] == 4}, y, {x, 0, 5}];
sol2 = NDSolve[{Derivative[1][y][x] == -y[x]^3 + y[x]^2 + y[x]*\[Lambda] - \[Lambda] /.
\[Lambda] -> myLambda, y[0] == 0.82}, y, {x, 0, 5}];
sol3 = NDSolve[{Derivative[1][y][x] == -y[x]^3 + y[x]^2 + y[x]*\[Lambda] - \[Lambda] /.
\[Lambda] -> myLambda, y[0] == -4}, y, {x, 0, 5}];
sol4 = NDSolve[{Derivative[1][y][x] == -y[x]^3 + y[x]^2 + y[x]*\[Lambda] - \[Lambda] /.
\[Lambda] -> myLambda, y[0] == 1.1}, y, {x, 0, 5}];
Show[{Plot[Evaluate[y[x] /. First[sol1]], {x, 0, 5},
PlotRange -> {{0, 5}, {-5, 5}}], Plot[Evaluate[y[x] /. First[sol2]],
{x, 0, 5}, PlotRange -> {{0, 5}, {-5, 5}}],
Plot[Evaluate[y[x] /. First[sol3]], {x, 0, 5},
PlotRange -> {{0, 5}, {-5, 5}}], Plot[Evaluate[y[x] /. First[sol4]],
{x, 0, 5}, PlotRange -> {{0, 5}, {-5, 5}}],
Graphics[{Dashed, Line[{{-5, 1}, {5, 1}}]}],
Graphics[{Dashed, Line[{{-5, Sqrt[myLambda]}, {5, Sqrt[myLambda]}}]}],
Graphics[{Dashed, Line[{{-5, -Sqrt[myLambda]}, {5, -Sqrt[myLambda]}}]}]}]