Let $l_1, l_2, r_1, r_2 $ be negative real numbers such that $r_1r_2 \neq 0$ and consider the two matrices $$L = \left(\begin{matrix} l_1&0\\ 0& l_2 \end{matrix} \right),\,\, R= \left(\begin{matrix} r_1&0\\ 0& r_2 \end{matrix} \right).$$
Let now $g: \mathbb{R}^2 \to \mathbb{R}^2, g(X) = |X|^2 X,$ where $|.|$ is the Euclidean norm in $\mathbb{R}^2.$
I'd like to find a solution $f:\mathbb{R}^2 \to \mathbb{R}^2$ to the differential equation
$$Df(X) LX + Rf(X) = g(X),$$
where $Df(X)$ denotes the Frechet derivative at the point $X.$
The idea in this setting is to introduce $t\mapsto Y(t)$ verifying
$$\frac{d}{dt}Y = LY, \,\, Y(0) = X.$$
The well known candidate is $Y(t) = e^{tL}X.$
Hence we can write
$$\partial_t \left(e^{tR}g(Y)\right) = e^{tR} Dg(Y)LY + Re^{tR}g(Y).$$
Integrating over $(0, +\infty)$ the above equality and denote $f(X) = - \int_0^\infty e^{tR}g(e^{tL}X)dt$, we find that
$$g(X) =g(Y(0)) - \lim_{t\to +\infty}e^{tR}g(Y(t)) = Df(X)LX+Rf(X).$$
Remark: The above integral defining $f$ is convergent since
$$| e^{tR}g(e^{tL}X)| \leq e^{t \max(r_1, r_2)}|X|^3.$$
My question is: Is there a similar method to give a solution to the differential equation
$$Df(X) \tilde{L}X + \tilde{R}f(X) = g(X),$$ where
$$ \tilde{L} = \left(\begin{matrix} l_1&0\\ 0& -l_2 \end{matrix} \right),\,\,
\tilde{R}= \left(\begin{matrix} r_1&0\\ 0& -r_2 \end{matrix} \right).$$
Thank you for any hint.