I have recently been trying to find an exact solution to the following system of first-order ODEs:
$\begin{cases} \frac{dx}{dt}=y(t)*z(t) \\ \frac{dy}{dt}=x(t)*z(t) \\ \frac{dz}{dt}=x(t)*y(t)\end{cases}$
Does this system even have, for a given initial condition involving $x,y,z,$ and $t$, a non-trivial solution in terms of elementary functions? I request guidance in finding these non-trivial solutions.
Let's introduce the auxiliary variable $\tau(t)$ defined by the differential equation $$ \frac{d\tau}{dt}=yz,\qquad \tau(0)=0. \tag{1} $$ Then, assuming $yz\neq 0$, we have $$ \frac{dx}{d\tau}=\frac{\frac{dx}{dt}}{\frac{d\tau}{dt}}=1 \implies x(\tau)=x_0+\tau, \tag{2} $$ $$ \frac{dy}{d\tau}=\frac{\frac{dy}{dt}}{\frac{d\tau}{dt}}=\frac{x}{y}=\frac{x_0+\tau}{y} \implies \frac{y^2}{2}=\frac{y_0^2}{2}+x_0\tau+\frac{\tau^2}{2} $$ $$ \implies y(\tau)=\text{sgn}(y_0)\sqrt{y_0^2+2x_0\tau+\tau^2}, \tag{3} $$ and, similarly, $$ \frac{dz}{d\tau}=\frac{\frac{dz}{dt}}{\frac{d\tau}{dt}}=\frac{x}{z}=\frac{x_0+\tau}{z}\implies z(\tau)=\text{sgn}(z_0)\sqrt{z_0^2+2x_0\tau+\tau^2}. \tag{4} $$ Finally, plugging $(3)$ and $(4)$ into $(1)$ and solving for $t$, we obtain $$ t(\tau)=\text{sgn}(y_0z_0)\int_0^{\tau}\frac{ds}{\sqrt{(y_0^2+2x_0s+s^2)(z_0^2+2x_0s+s^2)}}. \tag{5} $$ Equations $(2)$-$(5)$ are the solution, in parametric form, to the given system of ODEs$^{(*)}$.
$^{(*)}$ Some of the equations above don't make sense if one or more of $x_0, y_0, z_0$ is zero. These cases must be treated separately.