We want to find a function $F(x(t),y(t))=c$ where $x(t),y(t)$ are solutions to the system $\begin{bmatrix} \dot x=\frac{t-y}{y-x} \\\dot y=\frac{x-t}{y-x}\end{bmatrix}$. Such a function $F(x(t),y(t))$ is called the integral of the system.
I seem to run into difficulties, this is what i did:
Divide the second row by the first row to get: $\frac{ \dot y}{\dot x} = \frac{dy}{dx}=\frac{x-t}{t-y}$ and from this we can see that $\int t-ydy=\int x-tdx$ or in other words $yt-\frac{1}{2}y^2=\frac{1}{2}x^2-xt+c$, move everything to one side to get $yt-\frac{1}{2}y^2-\frac{1}{2}x^2+xt=c$
The problem is that $t$ shouldn't be there. the $F(x,y)$ we are looking for is a function of $x,y$ only. It's true that $x,y$ are functions of $t$ but still $F(x,y)$ should be an expression without $t$. What have I done wrong?
Edit: Another approach is to add both rows of the system to get $\dot x +\dot y =-1$, integrate with respect to $t$ to get $x+y=-t+c$, still not very helpful, $t$ is still there.
i will start with the differential equation $$\frac{dx}{dy} = \frac{t-y}{x-t}$$ this can be rearranged to give $$x\, dx + y\, dy=t(dy + dx) =t\, dt\left(\frac{dx}{dt} + \frac{dy}{dt} \right)=t\, dt\left( \frac{t-y}{y-x}+\frac{x-t}{y-x}\right) = -tdt$$ that is $$x^2 + y^2 + t^2 = constant $$
if we have the initial conditions $$x = x_0, y= y_0 \text{ at time } t = 0, \text{ then } x^2 + y^2 = x_0^2 + y_0^2 - t^2$$ i have not figured out another constrain/constant of motion so that we can solve for both $x$ and $y$.