I would like to know if the following differential equation ($\alpha,\beta,\gamma,d,\Lambda,w$ are constants)
$x'''(t)=\frac{1}{24 (3 \alpha -\beta )}\frac{x(t)^{-3 w-2}}{x'(t)} \left(36 \alpha x(t)^{3 w+2} x''(t)^2-12 \beta x(t)^{3 w+2} x''(t)^2+108 \alpha x(t)^{3 w} x'(t)^4-36 \beta x(t)^{3 w} x'(t)^4-6 \gamma x(t)^{3 w+2} x'(t)^2-72 \alpha x(t)^{3 w+1} x'(t)^2 x''(t)+24 \beta x(t)^{3 w+1} x'(t)^2 x''(t)+d x(t)+2 \gamma \Lambda x(t)^{3 w+4}\right)$
where prime denotes derivative with respect to $t$ of the function $x(t)$, can be linearized in some way in order to obtain information about the stability of the possible solutions. I have tried the standard approach of calling $x'(t)=A$, $x''(t)=B$ and expanding in Taylor series, but I can't get rid of the nonlinearities.
Hint.
After transforming to the canonical form
$$ \cases{\dot x_1 = x_2\\ \dot x_2 = x_3\\ \dot x_3 = f(x_1.x_2,x_3)} $$
you can follow doing $\delta_i = x_i-x^0_i$ and
$$ \cases{ \dot \delta_1 = \delta_2\\ \dot \delta_2 = \delta_3\\ \dot \delta_3 = f(x^0_1,x^0_2,x^0_3)+\sum_k \frac{\partial f}{\partial x_k}|_{(x^0_1,x^0_2,x^0_3)}\delta_k } $$
Here $x^0_k = x^0_k(t)$
Example
$$ x'''(t)=\frac{\left(c_1x''(t)^2+c_2 x(t)^{-2} x'(t)^4\right)}{x'(t)} $$
in the canonical form it gives
$$ \cases{ \dot x_1 = x_2\\ \dot x_2 = x_3\\ \dot x_3 = \frac{\left(c_1x_3(t)^2+c_2 x_1(t)^{-2} x_2(t)^4\right)}{x_2(t)} } $$
then
$$ \cases{ \dot \delta_1 = \delta_2\\ \dot \delta_2 = \delta_3\\ \dot \delta_3 = \frac{\left(c_1x_3^0(t)^2+c_2 x_1^0(t)^{-2} x_2^ 0(t)^4\right)}{x_2^0(t)} -\frac{c_2x_2^0(t)^3}{x_1^0(t)^2}\delta_1 + \left(\frac{3c_2x_2^0(t)^2}{x_1^0(t)}-\frac{c_1x_3^0(t)^2}{x_2^0(t)^2}\right)\delta_2+\frac{2c_1x_3^0(t)}{x_2^0(t)}\delta_3 } $$
The functions $x_k^0(t)$ are all known so the resulting system is linear in $\delta_k$