I had a system of differential equations and I ended up in this.
$4x+2y+4z=0$ $2x+y+2z=0$ $4x+2y+4z=0$
I don't know what to do from here.
I had a system of differential equations and I ended up in this.
$4x+2y+4z=0$ $2x+y+2z=0$ $4x+2y+4z=0$
I don't know what to do from here.
Note that your system of equations is actually a single equation in various avatars. (i.e.) one is scalar multiple of the other. Put another way, these equations convey the same information, nothing new!
$4x+2y+4z=0 \iff 2(2x+y+2z=0) \iff 4x+2y+4z=0$ Since there are three unknowns, and only one equations, two parameters are required to fix a solution. We can have a parametric solution as follows:
Set $x=\lambda$ and $y=\mu$. Then, $2\lambda+\mu+2z=0 \implies z=\dfrac{-2\lambda-\mu}{2}$
Hence, the triplet $(\lambda,\mu,\dfrac{-2\lambda-\mu}{2})$ with $\lambda,\mu \in \mathbb R$ is a solution to your system.