In the paper "Negative Autoregulation Speeds the Response Times of Transcription Networks" (see https://doi.org/10.1016/S0022-2836(02)00994-4) they present an ODE describing negative autoregulation, but I do not understand how they solve it:
This is the ODE: $$ \frac{dx(t)}{dt}=\frac{\beta}{1+ \frac{x(t)}{k}} - \alpha x(t) $$ In the paper they claim that they use the steady state solution $$ x_{\text {st }}=\frac{\sqrt{k^2+4 k \frac{\beta}{\alpha}}-k}{2} $$ to integrate the ODE and obtain $$ \begin{aligned} t(x)-t\left(x_0\right)= & -\frac{1}{2 \alpha}\left[\log \left(\left(x-x_{\mathrm{st}}\right)\left(x+k+x_{\mathrm{st}}\right)\right)\right. \\ & \left.+\frac{k}{k+2 x_{\mathrm{st}}} \log \left(\frac{x-x_{\mathrm{st}}}{x+k+x_{\mathrm{st}}}\right)\right]_{x_0}^x \end{aligned} $$ which one can approximate with $$ \frac{x(t)}{x_{s t}}=\sqrt{1-\mathrm{e}^{-2 \alpha t}} $$ if $\frac{\beta}{\alpha}>>k$ where we can approximate the steady state with $x_{st} \approx \sqrt{\frac{\beta}{\alpha}k}$
For reference here is the solution according to Wolfram Alpha: https://www.wolframalpha.com/input?i=dx+%2F+dt+%3D+b+*+%28k+%2F+%28k+%2B+x%29%29+-+a+x
Also let me know how I can improve the question.
By the assumptions at the bottom, $k$ is a very small number so that one can approximately write for $x\gg k$ $$ \dot x = \frac{k\beta}{x}-\alpha x $$ This is a Bernoulli equation. Setting $z=x^2$ gives a linear equation $$ \dot z=2(k\beta-\alpha z)\implies k\beta-\alpha z(t) = ce^{-2\alpha t} $$ At infinity, $k\beta-\alpha x_{st}^2 = 0$ or $x_{st}=\sqrt{\frac{k\beta}{\alpha}}$, so that, with a different factor $c$ $$ x(t)^2=x_{st}^2\,(1-ce^{-2\alpha t}). $$ This is compatible with the size order assumption as $x\approx x_{st}\approx \sqrt{\frac\beta\alpha k}\gg \sqrt{k\,k}=k$
This is essentially your claim, only that there is an additional parameter to account for initial conditions. With $c=1$ the exact same formula results, satisfying $x(0)=0$.
However, then the magnitude condition $k\ll x$ is not satisfied for some initial segment. The approximation formula gives $x_1(t)\approx\sqrt{2\beta k t}$ for $t\approx 0$, which has a vertical tangent at $t=0$. But going back to the original equation you get $\dot x\approx \beta$ for $x\approx 0$, so $x_2(t)\approx\beta t$ has linear growth.
One can construct a cross-over point where $x_1(t_1)=x_2(t_2)$ and $\dot x_1(t_1)=\dot x_2(t_2)$. This gives $t_2=2t_1$ from considering the logarithmic derivative $\dot x/x$ and then $t_2=\frac k\beta$ for the equality of the values at $x=k$.
This means that to any time computed with the approximation formula, one should add a correction of $\frac{k}{2\beta}$ to account for these effects at $x=0$. One may consider this correction as too small to care ...