I'm solving the ODE \begin{align} \frac{\mathrm{d}^2x}{\mathrm{d}t^2}+\beta\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\omega_0^2x=0, \end{align} with initial conditions $x(t_0)=x_0$ and $\dot{x}(t_0)=\dot{x}_0=v_0$.
I start with the substitution \begin{align} \frac{\mathrm{d}x}{\mathrm{d}t}&=v\\ \frac{\mathrm{d}^2x}{\mathrm{d}t^2}&=v\frac{\mathrm{d}v}{\mathrm{d}x}, \end{align} so the ode becomes \begin{align} v\frac{\mathrm{d}v}{\mathrm{d}x}+\beta v^2+\omega_0^2x=0. \end{align} Using an integrating factor of $2e^{2\beta x}$ I arrive at \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}(v^2e^{2\beta x})+2\omega_0^2xe^{2\beta x}=0. \end{align} I've used definite integrals, which makes the constants messier but is where my question lies, \begin{align} \int_{x_0}^x\left(\frac{\mathrm{d}}{\mathrm{d}x}(v^2e^{2\beta x})+2\omega_0^2xe^{2\beta x}\right)\mathrm{d}x=\int_{x_0}^x0\mathrm{d}x, \end{align} yielding \begin{align} v^2e^{2\beta x}+\frac{\omega_0^2}{\beta}xe^{2\beta x}-\frac{\omega_0^2}{2\beta^2}e^{2\beta x}=v_0^2e^{2\beta x_0}+\frac{\omega_0^2}{\beta}x_0e^{2\beta x_0}-\frac{\omega_0^2}{2\beta^2}e^{2\beta x_0}. \end{align} Multiplying by $e^{2\beta x}$ and solving for $v^2$ gives \begin{align} v^2=\left(\frac{\omega_0^2}{2\beta^2}-\frac{\omega_0^2}{\beta}x\right)+\left(v_0^2+\frac{\omega_0^2}{\beta}x_0-\frac{\omega_0^2}{2\beta^2}\right)e^{2\beta(x_0-x)}=\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2. \end{align} This is quite messy but it is separable. That being the case I can't integrate it, and neither can <integral-calculator.com>.
That being so, if the term in front of the exponential term is zero, it is an easy solve. Simply requiring that the initial conditions satisfy that reduces this to a particular solution, but can't I just make the transformation from $x\rightarrow x'$ such that \begin{align} v_0'^{2}+\frac{\omega_0^2}{\beta}x'_0-\frac{\omega_0^2}{2\beta^2}=0? \end{align} This would simply be a translation of $x$, as $x'_0$ is still a constant.
I'm asking because it seems too easy to be true, which in my experience of solving ODE's always has meant I was making an error. As well as the only way of making this transformation that I can think of is by letting \begin{align} v'^{2}+\frac{\omega_0^2}{\beta}x'-\frac{\omega_0^2}{2\beta^2}=0, \end{align} which is obviously not a valid assumption, as it necessitates $x$ be a certain function of $t$. So what I'm after is what I'm doing wrong and why I can't do that, or how to go about this transformation if it is indeed 'math legal'.
Edit:
Realized that I just need make the transformation $x=x'+A$, then solve for an $A$ that satisfies the reduction that I want. My question about this being seemingly too good still stands.