Simple Differential Equation 
We have
  $$\frac{dx}{dt}=x-y-1$$
  $$\frac{dy}{dt}=x-y+1$$
Express $y$ in terms of only $x$ (i.e. no $t$ term).

My professor gave me the hint "use $\frac{d}{dt}(x-y)$", but I don't know how this is supposed to help me.
 A: Use both $x-y$ and $x+y$.  I get
$$\frac{d}{dt}(y-x) = 2 \implies y-x=2 t+C$$
$$\frac{d}{dt}(y+x) = 2 (x-y) = -4 t \implies y+x=-2 t^2 +C'$$
Do some algebra to find that
$$t^2+t+x=0 \implies t=-\frac12 \pm \frac12 \sqrt{1-4 x}$$
Then
$$y=t-t^2+C = x-1\pm\sqrt{1-4 x}+C$$
A: We can also do the problem with a long way. However, @Ron's approach is a shortcut concrete one, I 'd like to bring up additional points. They are useful to know. :)
The system is equal to the following system if we set $x'=Dx, y'=Dy$:
$$ \left\{
        \begin{array}{ll}
            (D-1)x+y=-1 \\
            -x+(D+1)y=1
        \end{array}
    \right.
$$ Now, let's try to solve it as it is solved in pre-calculus. We have:
$$ \left\{
        \begin{array}{ll}
            \color{green}{(+1)}(D-1)x+\color{green}{(+1)}y=\color{green}{(+1)}\times-1 \\
            \color{green}{(D-1)}(-x)+\color{green}{(D-1)}(D+1)y=\color{green}{(D-1)}\times1
        \end{array}
    \right.
$$ So; $$[(D^2-1)+1]y=-2\Longleftrightarrow D^2y=-2\Longleftrightarrow y=-t^2+C_1t+C_2$$ wherein $C_1$ and $C_2$ are constants. Doing the same way for $x$ you can find it with respect to $t$ also.
