Let's assume that we have the following system of O.D.E. \begin{align} x' = f_1(x,y,z,v),\ y' = f_2(x,y,z,v),\ z' = f_3(x,y,z,v),\ v' = f_4(x,y,z,v) \end{align} for suitable smooth functions $f_1$, $f_2$, $f_3$ and $f_4$.
Let $\lambda$ be some nonzero constant and let's assume that \begin{align} \lambda f_3(x,y,z,\lambda z) = f_4(x,y,z,\lambda z). \end{align} If one considers the `reduced' system of 0.D.E. \begin{align} x' &= f_1(x,y,z,\lambda z)\ y' = f_2(x,y,z,\lambda z)\ z' = f_3(x,y,z,\lambda z) \end{align} then, any solution of the reduced system is a solution of the original system provided that $v=\lambda z$.
My question is the following. From the implicit function theorem (applied to the equation $\lambda f_3 = f_4$, as before) we know that \begin{align} z = g(x,y), \end{align} for some function $g$.
If we consider the even smaller system of O.D.E. \begin{align} x' = f_1(x,y,g(x,y),\lambda g(x,y))\ y' = f_2(x,y,g(x,y),\lambda g(x,y)) \end{align} will any solution of the smallest system be a solution of the original system provided that $z = g(x,y)$ and $v = \lambda g(x,y)$?
Thank you so much for your help.
Yes! It will be a solution. Wherever you can find an First Integral to your systems, it can be reduced in number of variables. If your system admits enough first integrals, it is called an Integrable system .