I'm not sure if this is what you were looking for, but hopefully it helps.

[Fantini]

Consider the system of differential equations

$$x' = 2x - e^y (2+y),$$ $$y' = -y.$$

Find the stable and unstable manifolds near the rest point.

I know that the stable manifold $W^s$ is a immersed surface in $\mathbb{R}^2$ with tangent space $E^s$ (the stable linear subspace). How can I compute the stable and unstable manifolds? I can't find decent instructions anywhere. Thanks.

 

[I like Serena]

Consider the system of differential equations

$$x' = 2x - e^y (2+y),$$ $$y' = -y.$$

Find the stable and unstable manifolds near the rest point.

I know that the stable manifold $W^s$ is a immersed surface in $\mathbb{R}^2$ with tangent space $E^s$ (the stable linear subspace). How can I compute the stable and unstable manifolds? I can't find decent instructions anywhere. Thanks.

Let's start with a qualitative approach to the system of differential equations.
The vector plot looks like this.
(Just learned how to make vector plots with W|A. ;))

You're looking for a curve through the rest point such that its tangent satisfies the equations at all points (or is a multiple thereof).

So what is the rest point?
Can you solve the system and find a curve that includes the rest point?

 

[Fantini]

The rest point is found by solving simultaneously the equations. We find that $y=0$ and then $2x - 2 = 0$ or $x=1$. Therefore the rest point is $(1,0)$.

We also know that the linearization of the system is $$\begin{bmatrix} 2 & 0 \\ 0 & -1 \end{bmatrix}.$$ I believe we can conclude that the stable manifold has dimension one (it is a curve).

I've seen in some places that we assume that $y$ can be parametrized as a function of $x$ to get this curve, or $y = \phi (x)$. But then what?

 

[I like Serena]

Good. :)

I've seen in some places that we assume that $y$ can be parametrized as a function of $x$ to get this curve, or $y = \phi (x)$. But then what?

So... can you solve the system and write it as $y = \phi (x)$?
After that we'll see...

 

[Fantini]

What do you mean by solve the system? By setting each equation to zero or as they are? In the latter I don't think such is possible. This is why I read that we assume it can be written and work from there.

 

[I like Serena]

$$\begin{array}{ll}x' = 2x - e^y (2+y) & \qquad (1) \\
y' = -y & \qquad (2) \end{array}$$

Find the stable and unstable manifolds near the rest point.

I know that the stable manifold $W^s$ is a immersed surface in $\mathbb{R}^2$ with tangent space $E^s$ (the stable linear subspace). How can I compute the stable and unstable manifolds? I can't find decent instructions anywhere. Thanks.

What do you mean by solve the system? By setting each equation to zero or as they are? In the latter I don't think such is possible. This is why I read that we assume it can be written and work from there.

Let's see...
For example:
$$\begin{array}{lcll}
y' &=& -y & \qquad \text{starting from (2)}\\
\frac{y'}{y} &=& -1 \\
\ln y &=& -t + C \\
y &=& c_2 e^{-t} & \qquad (3)\\
t &=& -\ln(y/c_2) & \qquad (4)\\
\\
x' &=& 2x - e^y (2+y) & \qquad \text{continuing from (1)} \\
x' &=& 2x - e^{c_2 e^{-t}}(2+c_2 e^{-t}) & \qquad \text{substituting (3)}\\
\dots \\
x &=& c_1 e^{2t} + e^{c_2 e^{-t}} & \qquad (5)\\
x &=& c_1 e^{2 \cdot -\ln(y/c_2)} + e^y & \qquad \text{substituting (3) and (4)}\\
\dots
\end{array}
$$