Domain of a solution of differential equation Given the following system of differential equations
\begin{equation}
\begin{cases}
\frac{dx}{dt}=-x+xy
\\\frac{dy}{dt}=-2y-x^2
\end{cases}
\end{equation}
I want to prove that the maximal solutions of the system are defined on all $\mathbb{R}$ and that
$$\lim_{t\rightarrow +\infty } {(x(t),y(t))}=0$$
As shown here I tried proceeding same way but I don't have any information about $t=0$.
$$x′x+y′y=-x^2-2y^2\leq 0$$
$$x′x+y′y= x'+y'\cdot x+y =\frac{1}{2}\frac{d}{dt}|| x+y ||^2$$
$$\frac{1}{2}\frac{d}{dt}|| x+y ||^2\leq 0$$
Hence I have information about the monotony, but I don't know how to show that, for example, the solution is limited on a subset of $\mathbb{R}$ (I could use this hypothesis to prolong the solution on all $\mathbb{R}$).
 A: There are many ways to approach such things, which can be studied on a Dynamical Systems course.
(1) With some brute force, you can prove by solving each equation separately :
$$x(t) = c_1e^{\int [y(t)-1]dt}$$
$$y(t) = \frac{-\int e^{2t}x^2(t)dt+c_2}{e^{2t}}$$
and then make conclusions.
(2) We have the system : 
$$\begin{equation}
\begin{cases}
x'=-x+xy
\\y'=-2y-x^2
\end{cases}
\end{equation}$$
We observe that $O(0,0)$ is a stationary point for the system.
The Jacobian of this system, is : 
$$J(x,y) = \begin{bmatrix} -1+y & x \\ -2x & -2 \end{bmatrix}$$
and thus, the Jacobian of regarding the stationary point $O(0,0)$ is : 
$$J(0,0)=\begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}$$
Calculating the eigenvalues of the $J(0,0)$ :
$$\det(J(0,0)-λI)=0\Rightarrow(-1-λ)(-2-λ)=0 \Leftrightarrow \begin{cases} λ_1=-1 \\ λ_2 = -2 \end{cases}$$
Thus, it is : $λ_2 < λ_1 < 0$ and $λ_1\cdot λ_2 > 0$, which means $O(0,0)$ is an asymptotically stable node.
From the image of the phase plane that corresponds to an asymptotically stable node (check here), you can conclude that the solutions "collapse" to $O(0,0)$, thus : $\lim_{t\rightarrow +\infty } {(x(t),y(t))}=0$
(3) As mentioned in one of the comments, the first expression you have written (the inequality), shows that the functional $V(x,y) = x^2 + y^2$ is a strong Lyapunov one, so for $t \to + \infty$ you can make a conlcusion from Lyapunov's theorem. Though, as mentioned again, that needs an extra argument for $t<0$.
(4) Via Linearization with polar coordinates, you can transform your system as : 
$$x=r\cosθ$$
$$y=r\sinθ$$
$$x^2 + y^2 = r^2$$
Differentiating the last equation, yields : 
$$rr' = xx' + yy'$$
To find the angle $θ$, we take : 
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule, we get :
$$\theta' = \dfrac{x y' - y x'}{r^2}$$
Making substitutions you can thus yield results for a new system :
$$\begin{cases} r' = \dots \\ θ' = \dots \end{cases}$$
and make conclusions.
Regarding polar coordinates, you can check an answer of mine, here.
A: Note that you already have
$$x′x+y′y=-x^2-2y^2$$
which implies
$$ \frac{1}{2}(x^2+y^2)'=-x^2-2y^2\le-x^2-y^2=-(x^2+y^2). $$
So
$$ (x^2+y^2)'+2(x^2+y^2)\le0$$
or
$$ e^{2t}(x^2+y^2)'+2e^{2t}(x^2+y^2)\le0$$
or
$$ [e^{2x}(x^2+y^2)]'\le0. $$
Integrating from $0$ to $t$ gives
$$ e^{2t}(x^2(t)+y^2(t))\le x^2(0)+y^2(0) $$
or
$$ x^2(t)+y^2(t)\le e^{-2t}[x^2(0)+y^2(0)]. $$
So
$$ \lim_{t\to\infty}(x^2(t)+y^2(t))=0. $$
