Simple Differential Equation

Question

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.

Answer 1

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

$$y=t-t^2+C = x-1\pm\sqrt{1-4 x}+C$$

Answer 2

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.