I'm trying to solve this first order difference equation. I usually go through the usual avenue (eigenvalues of homogeneous function then eigenvectors then solution) but here I don't know what I can do to get $y$ and $x$ power $1$ so that it becomes a system of two equations and two unknowns.
$x_{t+1}=-x_{t}+2x_{t}^{2}$
$y_{t+1}=-2x_{t}^{2}-y_{t}$
This isn't a solution, just some ideas i threw around which resulted in interesting results
We have that:
$$ x_{t+1} + y_{t+1} = -x_t - y_t $$
Let $Df = f_{x+1} - f_{x} $
$$ Dx + Dy = -2(x+y)$$
$$ u = x+y$$
(by linearity)
$$ Du = -2u$$
Assume that $u = a^t$ for some a. We find the a by noting
$$ a^{x+1} - a^t = -2a^t \rightarrow a - 1 = -2 \rightarrow a = -1 $$
Thus it follows that
$$ x+y = C_1 (-1)^t$$
Now for the second equation we can observe a very close solution. Recall
$$ x_{t+1} = - x_t + 2 x_t^2 $$
It follows then that
$$ Dx = 2x(x-1) $$
Consider the simpler
$$ Dx = x(x-1) $$
If we assume $x = 2^{g(x)}$ then
$$ Dx = 2^{g(x+1)} - 2^{g(x)} = 2^{g(x)}(2^{Dg}-1) $$
if $Dg = g \rightarrow g = C_1 2^t$ thus we have that
$$ x = C_1 2^{C_2 2^{t}} $$
So something closely related to that is what you seek.