Take the following nonlinear differential equations $$ A_1 \ddot x_1(t) + B \dot x_1(t) + C x_1(t) = f(x_1) \tag{1} $$ and $$ A_2 \ddot x_2(t) + B \dot x_2(t) + C x_2(t) = f(x_2) \tag{2} $$ with $$ x_1(t=0) = x_2(t=0) = x_0\\ \dot x_1(t=0) = \dot x_2(t=0) =0. $$ Here, $B$, $C$ and $f(.)$ are equal for both equations. I simulated different examples for function $f(.)$ such as $sin(.)$, $exp(.)$, $\sqrt{.}$ and different combinations of nonlinear operations, and it always holds $$ x_1(t) = x_2(\sqrt{A_2/A_1}t). \tag{3} $$ Please note that $x_1(t)$ and $x_2(t)$ are obtained via numerical methods. In other word, due to the unknown nature of $f(.)$, no analytical solution is available.
Question When $f(x) = 0$, $(1)$ and $(2)$ will become homogeneous linear differential equations, for which $(3)$ is easily explainable. How can I explain $(3)$ when we are dealing with an unknown function of $f(x)$?
Introduce the function $g:x\mapsto C x - f(x)$ so that the differential equations read $$ A_i \ddot x_i + B\dot x_i + g(x_i) = 0 , \qquad i\in \lbrace 1,2\rbrace . $$ Now, let us substitute $x_1(t) = x_2(\tau)$ with $\tau = \alpha t$ and $\alpha = \sqrt{A_2/A_1}$ in the differential equation for $i=1$. By using the chain rule, we have $\dot x_1(t) = \alpha \dot x_2(\tau)$ and $\ddot x_1(t) = \alpha^2 \ddot x_2(\tau)$. Thus, by using the other differential equation ($i=2$), we have \begin{aligned} A_1 \ddot x_1(t) + B \dot x_1(t) + g(x_1(t)) &= A_1\alpha^2 \ddot x_2(\tau) + B\alpha \dot x_2(\tau) + g(x_2(\tau)) \\ &= A_2 \ddot x_2(\tau) + B\alpha \dot x_2(\tau) + g(x_2(\tau)) \\ &= B(\alpha-1) \dot x_2(\tau) \neq 0 \end{aligned} in general. Therefore, your claim is wrong for non-constant solutions unless $B=0$ or $A_1=A_2$.