Special way to solve systems of ODEs

Question

If we have a system of two differential equations and it can be written in the form: $$\frac{dx}{f_1(x,y,t)}=\frac{dy}{f_2(x,y,t)}=\frac{dt}{f_3(x,y,t)}=\gamma$$ Then we can conclude that: $$\frac{\alpha_1dx+\alpha_2dy+\alpha_3dt}{\alpha_1f_1(x,y,t)+\alpha_2f_2(x,y,t)+\alpha_3f_3(x,y,t)}=\gamma$$ Now if we choose the correct $\alpha_1,\alpha_2,\alpha_3$, Then we can get a solution. I don't understand if there is an easy way to get the correct coefficients of it is purely a guesswork. Here's an example: $$\frac{dx}{1+\sqrt{z-x-y}}=\frac{dy}{1}=\frac{dz}{2}=\gamma \Rightarrow \frac{\alpha_1dx+\alpha_2dy+\alpha_3dz}{\alpha_1(1+\sqrt{z-x-y})+\alpha_2+2\alpha_3}=\gamma$$ Then one easy tripple of coefficients is: $$\alpha_1=0 ;\\ \alpha_2=2 ;\\ \alpha_3=-1$$ Then after substitution we get: $$2dy-dz=d(2y-z)=0 \Rightarrow 2y-z=C_1$$ To get the full solution of the system, we need one more tripple of coefficients but I don't know what logic to use to find the most appropriate one. Can you help not only with this particular example but with this method as a whole? Also if you know how is it called because I can't even find it online to educate myself.

Answer 1

Take $\alpha_3 = 1$, $\alpha_1 = -1$, $\alpha_2 = -1$ and equate it with $dx/(1+\sqrt{z-x-y})$

you will get this type of equation, $dx/(1+\sqrt{(z-x-y))} = (dz-dx-dy)/\sqrt{(z-x-y)}$

Rewrite $(dz-dx-dy)$ as $d(z-x-y)$ and consider $t = z-x-y$. substitute it and solve it using variable separable method, integrate it and you will get your second answer.

Answer 2

It doesn't make sense to equate differential forms such as $\ \frac{dx}{f_1(x,y,t)},$$\frac{dy}{f_2(x,y,t)},$ or $\ \frac{dt}{f_3(x,y,t)}\ $ to a constant $\ \gamma\ $. Depending on context, $$ \frac{dx}{f_1(x,y,t)}=\frac{dy}{f_2(x,y,t)}=\frac{dt}{f_3(x,y,t)} $$ might be equivalent to the differential equations \begin{align} \frac{dx}{dt}&=\frac{f_1(x,y,t)}{f_3(x,y,t)}\\ \frac{dy}{dt}&=\frac{f_2(x,y,t)}{f_3(x,y,t)}\ , \end{align} where $\ x\ $ and $\ y\ $ are functions of $\ t\ $, or to \begin{align} \frac{dx}{ds}&=f_1(x,y,t)\\ \frac{dy}{ds}&=f_2(x,y,t)\\ \frac{dt}{ds}&=f_2(x,y,t)\ , \end{align} where $\ x,y\ $ and $\ t\ $ are functions of $\ s\ $. In this latter case, you could write $$ \frac{dx}{f_1(x,y,t)}=\frac{dy}{f_2(x,y,t)}=\frac{dt}{f_3(x,y,t)}=ds\ , $$ but I'm not aware of any context where $$ \frac{dx}{f_1(x,y,t)}=\frac{dy}{f_2(x,y,t)}=\frac{dt}{f_3(x,y,t)}=\gamma $$ makes sense when $\ \gamma\ $ is a fixed constant.