Finding the stable, unstable, center manifolds of a nonlinear system Question:
Consider the system
\begin{align}
\frac{dx}{dt} & = x-xy \\
\frac{dy}{dt} & = 2x-3y+y^2
\end{align}
Find the stable, unstable, center manifolds about origin $(0,0)$ up to and including cubic term.

Attempt:
So first, I computed the Jacobian matrix to be
$$J(x,y) = \begin{pmatrix} 1-y & -x \\ 2 & -3+2y \end{pmatrix}$$
Evaluated at the origin, this gives
$$J(0,0) = \begin{pmatrix} 1 & 0 \\ 2 & -3 \end{pmatrix}$$
and the eigenvalues are $1$ and $-3$, which means that the origin is a saddle (i.e. unstable).
The corresponding eigenvectors are $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
...
...
and I'm stuck T_T . What am I supposed to do from here?
Any hints would be much appreciated.
...please?
 A: Due to the structure of the eigenvectors, both having a non-zero $y$-component, you can take $y$ as parameter for both the stable manifold close to $x=2y$ and the unstable manifold close to $x=0$. Both thus are solutions of the (singular) ODE
$$
\frac{dx}{dy}=\frac{\dot x}{\dot y}=\frac{x\,(1-y)}{2x−3y+y^2}
$$
with condition $\lim_{y\to 0}x(y)=0$.
Unstable manifold for eigenvalue $\lambda=1$
Set $x=2y+yh(y)$, $h(y)=o(y)$, and insert into the ODE,
\begin{align}
2+(yh(y))'=\frac{dx}{dy}&=\frac{(2+h(y))(1-y)}{1+2h(y)+y},
\\[.5em]
(yh(y))'&=\frac{-4y-3h(y)-yh(y)}{1+2h(y)+y}.
\end{align}
Now determine the lowest degree term $h(y)=c_0y^r+...$, $r>0$, so that we get
$$
c_0(1+r)y^r=-4y-3c_0y^r+o(y^{1+r}).
$$
To get equal lowest degrees on both sides one needs $r=1$ and thus $5c_0=-4$. Inserting the power series $h(y)=y(c_0+c_1y+c_2y^2+...)$ the next coefficients are determined by the equations
\begin{align}
y(2c_0+3c_1y+4c_2y^2+...)(1+(2c_0+1)y+2c_1y^2+...)
&=-y(4+3c_0+(3c_1+c_0)y+(3c_2+c_1)y^2...)\\~\\
3c_1+2c_0(2c_0+1)&=-(3c_1+c_0)\\
4c_2+3c_1(2c_0+1)+4c_0c_1&=-(3c_2+c_1)\\
\end{align}
so that 
\begin{alignat}1
c_1&=-\frac16(3c_0+4c_0^2)&&=-\frac1{30}c_0(-15+16)&=-\frac{2}{75}, \\
c_2&=-\frac17(4c_1+10c_0c_1)&&=-\frac1{5⋅7}c_1(-20+40)&=-\frac{8}{21⋅25}. \end{alignat}
Combined this gives
$$
x(y)=2y-\frac45y^2-\frac{2}{75}y^3-\frac8{525}y^4+...
$$

Stable manifold for eigenvalue $λ=-3$
The equation for the stable manifold 
$$
\frac{dx}{dy}=\frac{x\,(1-y)}{2x−3y+y^2}
$$
has the solution $x(y)=0$ and it is the unique solution tangent to the $y$-axis.
