I'm trying to figure out how to solve this problem from "A Modern Introduction to Dynamical Systems" of R. J. Brown. The statement reads
For $r<0$ a constant, construct a second–order constant coefficient ODE whose time–1 map is $f(x)=rx$. (Hint: Work with the first-order system and then convert back at the end).
My attempt to solve the problem starts by knowing that every second–order constant coefficient ODE is given by $$ \ddot{x}(t) + a\dot{x}(t) + bx(t) = 0 $$ with $a,b\in\mathbb{R}$. We know that a possible solution for this equation (even if it isn't the general solution) is $$ x(t) = ce^{\lambda t} \qquad\text{with}\qquad \lambda = \frac{-a\pm\sqrt{a^2-4b}}{2} $$ and also considering the boundaries $$ x(0) = x_0 \qquad\text{and}\qquad x(1) = f(x_0) = rx_0 $$
I have that $$ x(0) = x_0 = c $$ and $$ x(1) = rx_0 = ce^{\lambda} $$ and that means that $$ \ln(r) = \frac{-a\pm\sqrt{a^2-4b}}{2} $$ but I don't know how to proceed from here.
There is also a solution online, that can be found here, but I don't understand how does he transforms $g:\mathbb{R}^2\to\mathbb{R}$ to $g:\mathbb{N}\times\mathbb{R}\to\mathbb{R}$ and just substitutes $n$ to $t$ to find the equation for $g$.
You have $\lambda=\ln(r)=\ln|r|+i\pi$, since $r<0$. Therefore, the second-order ODE is given by $$ (D-\lambda)(D-\bar{\lambda})x=0\implies (D^2-(\lambda+\bar{\lambda})D+\lambda\bar{\lambda}D)x=0, \tag{1} $$ where $D:=\frac{d}{dt}$. The real solutions to $(1)$ are given by $$ x(t)=e^{t\ln|r|}(A\cos(\pi t)+B\sin(\pi t))=|r|^t(A\cos(\pi t)+B\sin(\pi t)). \tag{2} $$ They satisfy $x(1)=-|r|x(0)=rx(0)$.