Please consider the following ODE:
$$\dot x = e^t \left( a x^2 + b x \right)$$
for which obviously $x=0$ is an equilibrium point. How can I find the conditions at which the origin is asymptotically stable? What should the Lyapunov function be?
What about an extended version?
$$\dot x = e^t \left( a x^3 + b x^2 + c x \right)$$
where the coefficients $a$, $b$, and $c$ are real.
On
$$\frac{dx}{dt} = e^t \left( a x^2 + b x \right)\quad\implies\quad e^t dt=\frac{dx}{a x^2 + b x}$$ $$e^t=\int \frac{dx}{a x^2 + bx }+c=\frac{1}{b}\ln\left|\frac{x}{ax+b}\right|+c$$ $$\boxed{x(t)=\frac{b}{a+C\:e^{-b\: e^t}}}\qquad C=\text{constant}$$
$$\frac{dx}{dt} = e^t \left( a x^3 + b x^2+cx \right)\quad\implies\quad e^t dt=\frac{dx}{a x^3 + b x^2+cx} \quad\implies\quad e^t=\int \frac{dx}{a x^3 + b x^2+cx}$$ $$e^t=\frac{1}{2c}\ln\left|\frac{ax^2+bx+c}{x^2} \right|+\frac{b}{c\sqrt{4ac-b^2}}\tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)+\text{constant}$$ The inverse function $x(t)$ cannot be written explicitly with a finite number of standard functions. One should be satisfied with the above implicit form of solution.
Given the system (1st case)
$$\dot{x} = e^{t}\left(a x^{2} + b x\right).$$
Let $a = - \alpha$ and $b = - \beta$, where $\alpha \geq 0$ and $\beta > 0$, then
$$\dot{x} = - e^{t}\left(\alpha x^{2} + \beta x\right).$$
If we choose the Lyapunov candidate function as
$$V = \frac{1}{2} x^{2},$$
then $\dot{V}$ becomes
$$\dot{V} = x \dot{x}$$
$$\dot{V} = - x^{2} e^{t} \left(\alpha x + \beta\right).$$
Since this term $x^{2} e^{t} > 0$ for $x \neq 0$ and $t > 0$, then in order for $\dot{V} < 0$, we must have $$\alpha x + \beta > 0$$ and therefore, the stability condition can be determined $$x > - \frac{\beta}{\alpha}$$
If we fix $\alpha = 1$, and the operating range of $x$ is known $\left[x_{\text{lb}}, x_{\text{ub}}\right]$, then $\beta$ can be designed such that it satisfies
$$x_{\text{lb}} > - \beta$$
You should be able to perform the analysis and do the same for the 2nd case.