First integrals for solving system of ODEs Assume a problem $$
\begin{cases}
\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{y}{x-y}, \\[2ex]
\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{x}{x-y}.
\end{cases}$$
Additionally, $x = x(t)$ and $y=y(t)$.
Attempt
Multiply the first equation by $x$ and the second one by $y$. Subtract the latter from the former.$$xx'-yy'=0$$
Integrate.
$$\frac{x^2}{2}-\frac{y^2}{2}=C_0$$
Next, let us simply substract the two original equations from one another.
$$\frac{\mathrm{d(x-y)}}{\mathrm{dt}}=-1$$
Multiply both sides by $\mathrm{dt}$ and integrate.
$$x-y=-t+C$$
Therefore, we have reached the solutions
$$
\begin{cases}
\frac{x^2}{2}-\frac{y^2}{2}=C_0, \\[2ex]
x-y=-t+C
\end{cases}\Rightarrow 
\begin{cases}
x^2-y^2+C_1=0=\Psi_1(x,y) + C_1, \\[2ex]
x-y+t+C_2=0=\Psi_2(x,y) + C_2.
\end{cases}$$
Question


*

*How can I show that $\Psi_1(x,y)$ and $\Psi_2(x,y)$ are both first integrals?

*How would one demonstrate that these are independent first integrals (assuming 1. stands)?

*Is going through points 1 and 2 sufficient to conclude that these form the general solution of the original system?


Don't forget $x^2-y^2=(x+y)(x-y)$,
\begin{align*}
  \frac{(x+y)(C-t)}{2} &= C_{0} \\
  x+y &= \frac{2C_{0}}{C-t}
\end{align*}
Hence,
$$\left \{
\begin{align*}
  x &= \frac{C_{0}}{C-t}+\frac{C-t}{2} \\
  y &= \frac{C_{0}}{C-t}+\frac{t-C}{2} \\
\end{align*}  \right.$$
