Existence of a solution of a system of ODEs Consider the following system of ODEs,
\begin{equation}
\begin{cases}
x'=-x^2-xy+x\\
y'=-y^2-xy+y
\end{cases}
\end{equation}
with initial conditions $x(0)=x_0>0$ and $y(0)=y_0>0$.
I should prove that $(x(t),y(t))$ is defined for all $t\geq 0$ and that $x(t)>0$ and $y(t)>0$ for all $t\geq 0$. Any suggestion?
 A: Assuming $x\ne0$ and $y\ne0$, dividing ($1$) by $x$ and ($2$) by $y$ gives
\begin{equation}
\dfrac{x^\prime}{x}=\dfrac{y^\prime}{y}
\end{equation}
thus
\begin{equation}
\dfrac{d}{dt}\ln(x)=\dfrac{d}{dt}\ln(y)
\end{equation}
thus 
\begin{equation}
x(t)=c_1y(t)
\end{equation}
So equation ($2$) becomes
\begin{equation}
y^\prime-y=-(1+c_1)y^2
\end{equation}
which is an easy Bernoulli equation.
Solving the Bernoulli equation gives
\begin{align}
x(t)&=\dfrac{c_1}{c_2e^{-t}+c_1+1}\\
y(t)&=\dfrac{1}{c_2e^{-t}+c_1+1}
\end{align}
So $x(0)=\dfrac{c_1}{c_1+c_2+1}>0$ and $y(0)=\dfrac{1}{c_1+c_2+1}>0$.
So we know $1+c_1+c_2>0$ and $c_1>0$ and that $x(t)>0,y(t)>0$ for all $t$. 
Proceeding upon an assumption that neither $x(t)$ nor $y(t)$ is ever $0$ we do obtain solutions for which that is true. 
This leaves open the question whether there is some other solution to the system for which that is false (other than the obvious $x(t)=y(t)=0$).
Note that 
\begin{equation}
y^\prime=f(x,y)=-y^2-xy+y
\end{equation}
is continuous on a neighborhood of every $(t_0,y_0)$ and that
\begin{equation}
\dfrac{\partial f}{\partial y}=-2y-x+1
\end{equation}
is also continuous on a neighborhood of every $(t_0,y_0)$ so there is a unique solution for equation ($2$) containing each $(t_0,y_0)$, in this particular case at $(0,y_0)$. The same argument goes for the equation ($1$). There is a unique equation for ($1$) containing $(0,x_0)$.
A: Assuming that $x(t),y(t)$ remain positive, we obtain, as in the answer of  John Wayland Bales, that $y(t)=cx(t)$, for some $c>0$ - In fact, $c=y_0/x_0$ - Also, $x$ satisfies the IVP
$$
x'=x-(1+c)x^2, \quad x(0)=x_0.
$$
The above IVP possesses a unique solution $\varphi(t)$, which is positive (not hard to show) and is defined for all $t>0$. Hence, the vector $\big(\varphi(t), c\varphi(t)\big)$ is a solution of the IVP
\begin{equation}
\begin{cases}
x'=-x^2-xy+x,\\
y'=-y^2-xy+y,
\end{cases} \qquad x(0)=x_0,\quad y(0)=y_0.
\end{equation}
As the above IVP enjoys global uniqueness (it possesses a Lipschitz flux), then $\big(\varphi(t), c\varphi(t)\big)$ is the unique solution, it is positive and defined for all $t\ge 0$.
Note also that
$$
\lim_{t\to\infty} x(t)=\frac{1}{1+c}=\frac{x_0}{x_0+y_0}, \quad
\lim_{t\to\infty} x(t)=\frac{y_0}{x_0+y_0}.
$$  
