For $x' = x^2$, and $y' = x + y$, Find all equilibrium points and decide whether they are stable, asymptotically stable, or unstable. For $x' = x^2$, and $y' = x + y$, Find all equilibrium points and decide whether they
are stable, asymptotically stable, or unstable.
I found that the equilibrium points are (0,0). After linearized the system, it become $x' = 0$ and $y' = x+y$ So, one eigenvalue is 0 and the other is 1 right?..Therefore, this is unstable.
 A: The equilibrium points of the system
$x' = x^2, \tag{1}$
$y' = x + y, \tag{2}$
occur where $x' = y' = 0$; by (1), we must have $x = 0$ at any such point, and by (2) it then follows tha $y = 0$ as well; $(0, 0)$ is the only equilibrium.  Calculating the Jacobean matrix $J$ of the system we see that
$J(x, y) = \begin{bmatrix} 2x & 0 \\ 1 & 1 \end{bmatrix}, \tag{3}$
whence at $(0, 0)$
$J(0, 0) = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}. \tag{4}$
The characteristic polynomial of $J(0, 0)$ is thus given by
$\det \begin{bmatrix} 0 - \lambda & 0 \\ 1 & 1 - \lambda \end{bmatrix} = \lambda(\lambda - 1), \tag{5}$
so we see the eigenvalues are $0$ and $1$.  The eigenvectors of $J(0, 0)$ are easily seen to be $(1, -1)^T$, corresponding to $\lambda = 0$, and $(0, 1)^T$ for $\lambda = 1$.  Since the eigenvalues do not have non-vanishing real parts, we cannot directly infer from the stable manifold theorem (look in up in wikipedia at wiki/Stable_manifold_theorem wiki/Stable_manifold_theorem, google it, or check out the discussion given in my answers to this question) that the local behavior of (1)-(2) is given directly by the eigenvalues and eigenvectors of $J(0, 0)$, but we can still use this information as a guide to our intuition in conducting further analyses.  The eigenvector $(0, 1)$ with $\lambda = 1$ is a strong indicator that the $y$-axis might exhibit special behavior; and sure enough, it is easily seen by inspection of (1)-(2) that there is a solution 
$x(t) = 0, \tag{6}$
$y(t) = y_0e^{(t - t_0)}, \; y(t_0) = y_0, \tag{7}$
for any $y_0 \in \Bbb R$.  Since $y(t) \to \text{sign}(y_0) \infty$ as $t \to \infty$, the system is unstable at $(0, 0)$.  Returning to (1)-(2), we see that $x$ is decoupled from the remainder of the system $y' = x + y$; thus we may solve for it independently of $y$.  We have, as long as $x \ne 0$,
$ (-x^{-1})' = x^{-2}x' = 1, \tag{8}$
so that
$\int_{x_0}^x x^{-2}dx = \int_{x_0}^x (-x^{-1})'dx = \int_{t_0}^t dt \tag{9}$
or
$-(\dfrac{1}{x} - \dfrac{1}{x_0}) = t - t_0, \tag{10}$
which after a little algebraic re-arranging yields
$x(t) = (x_0^{-1} - (t - t_0))^{-1} = \dfrac{x_0}{1 - x_0(t - t_0)}, \; x(t_0) = x_0 \ne 0; \tag{11}$
it can be seen from (11) that, for $x_0 > 0$, $x(t)$ grows with increasing $t$, and is in fact unbounded, increasing towards $\infty$ as $t \to t_0 + x_0^{-1}$ from below; as $t$ decreases, we have $x(t) \to 0$ from above as $t \to -\infty$.  For $x_0 < 0$, $x(t) \to 0$ from below as $t \to \infty$; as $t$ decreases, $x(t) \to -\infty$ as $t \to t_0 + x_0^{-1}$ from above.  In the region where $x(t)$ is defined, one can in fact use (11) in (2) to obtain
$y' = y + \dfrac{x_0}{1 - x_0(t - t_0)}, \tag{12}$
and solve (12) via the standard formulas (variation of parameters), yielding
$y(t) = y_0e^{(t - t_0)} + e^{(t - t_0)}\int_{t_0}^t \dfrac{x_0e^{-(s - t_0)}}{1 - x_0(s - t_0)}ds. \tag{13}$
The integral on the right-hand-side of (13) looks to me like it will blow up whenever $x(t)$ does, under the same conditions on $t$; I think it is pretty easy to show this is the case, but I will leave a proof of that hypothesis, and further analysis, to my readers.  I note however that, having at least one family of solutions $(0, y_0e^{(t - t_0)})$ which tends to $\text{sign}y_0 \infty$ as $t \to \infty$, the system (1)-(2) is unstable.  
The engaged reader may want to try an sketch a phase portrait for this system; I suspect that for $x > 0$ it looks like a repellor at $(0, 0)$, while in the region $x < 0$ it looks like a saddle.  Such behavior often occurs when the Jacobean has a zero eigenvalue.
Well, I have to get ready to go to work, so I can say no more at present.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
