<s>[INST] Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of equations in a neighborhood of the origin, $$x' = y+y^2 - 2xy + x^2$$ $$y'=x+y^2 - 2xy + x^2.$$ 
$\textbf{Solution:}$ Subtracting $x' = y+y^2 - 2xy + x^2$ from $y'=x+y^2 - 2xy + x^2$ gives us $y'-x' = x-y$ implies the following $$ x'+x=y'+y. \hspace{35pt} (1)$$ Integrating $x' = y+y^2 - 2xy + x^2$ with respect to $y$ and $y'=x+y^2 - 2xy + x^2$ with respect to $x$ gives us $$x = \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y \hspace{35pt} (2)$$ $$y=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3}. \hspace{35pt} (3)$$
Applying (2) and (3) to (1) gives us $$\frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y +y^2 -2xy+x^2$$ $$=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x + y^2 -2xy + x^2$$ $$\implies \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y = \frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x. \hspace{35pt}(4)$$
So, equation (4) denotes a function which if we replace $y$ with $x$ equation will be the same throughout. Thus, if a function of the form $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} - t^3 + t^2 + t \text{ where } t = x, y$$ implies $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} + t. \hspace{35pt} (5)$$
So equations (1), (4), and (5) define the stable and unstable points around the origin. So, $f'(x,y) >0$ as $$f'(x,y) = t + t^2 + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}.$$ Therefore, it will be unstable and we are done. 
 [/INST] $$x'=y+(x-y)^2 $$
$$y'=x+(x-y)^2$$ 
HINT :
$$y'-x'=x-y\quad\implies\quad \frac{y'-x'}{y-x}=-1\quad\implies\quad y-x=c_1e^{-t}$$
$$y=x+c_1e^{-t}$$
$y'=x+(c_1e^{-t})^2=x'-c_1e^{-t}$
$$x'-x=c_1e^{-t}+c_1^2e^{-2t}$$
$$x(t)=c_2e^t-\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$
$$y(t)=c_2e^t+\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$
This is the explicit solution $x(t)$ and $y(t)$ . 
I suppose that you can take it from here about stability.
Note : Eliminating $t$ from the above equations gives the trajectory equation $\quad 2(y-x)^3+3(y^2-x^2)=C \quad;\quad C=6c_1c_2$ .
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