First integrals for solving system of ODEs

Question

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

1. How can I show that $\Psi_1(x,y)$ and $\Psi_2(x,y)$ are both first integrals?
2. How would one demonstrate that these are independent first integrals (assuming 1. stands)?
3. Is going through points 1 and 2 sufficient to conclude that these form the general solution of the original system?

Answer 1

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, $$\left \{ \begin{align*} x &= \frac{C_{0}}{C-t}+\frac{C-t}{2} \\ y &= \frac{C_{0}}{C-t}+\frac{t-C}{2} \\ \end{align*} \right.$$