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 $y=y(t)$.


Multiply the first equation by $x$ and the second one by $y$. Subtract the latter from the former.$$xx'-yy'=0$$



Next, let us simply substract the two original equations from one another.


Multiply both sides by $\mathrm{dt}$ and integrate.


Therefore, we have reached the solutions $$ \begin{cases} \frac{x^2}{2}-\frac{y^2}{2}=C_0, \\[2ex] x-y=-t+C \end{cases}\Rightarrow \begin{cases} x^2-y^2+C_1=0=\Psi_1(x,y) + C_1, \\[2ex] x-y+t+C_2=0=\Psi_2(x,y) + C_2. \end{cases}$$


  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?
  • $\begingroup$ What are $\Psi_1$ and $\Psi_2$? $\endgroup$ – Chee Han Jun 7 '16 at 11:50
  • $\begingroup$ Sorry, just saw it. $\endgroup$ – Chee Han Jun 7 '16 at 11:50

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

  • $\begingroup$ Thanks for the answer! I am still a little confused on my question's points 1 and 2, but this gets me going in at least some direction ;) $\endgroup$ – Linear Christmas Jun 9 '16 at 12:55

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