Let $y''+f(t)y=0$ where $\lim\limits_{t\rightarrow\infty}f(t)=1$.
When is the critical point $(0,0)$ stable?
I want to show that this is the case when $\int_0^\infty|f(t')-1|dt'$ is finite.
For this we write $y(t)=A(t)\cos(t)+B(t)\sin(t)$ s.t. $A'\cos(t)+B'\sin(t)=0.$
I have shown that for $C(t)=A^2(t)+B^2(t)$, we have $$C'\leq4|f(t)-1|C.$$
And I want to use this to prove that $(0,0)$ is stable when $\int_0^\infty|f(t')-1|dt'$ is finite.
Plus, I know that stability means that for every neigbourhod $N$ of $(0,0)$ there is a $M\subset N$ s.t. if $y\in M$ then the flow $\phi(t,y)\in N$for all $t$.
What can I do next?
You can appoach this using Lyapunov theory.
Consider the states $x_1=y$, $x_2=\dot{y}$. If you define a Lyapunov function candidate $$V=x_1^2+x_2^2$$ then its derivative takes the form $$\dot{V}=2(1-f(t))x_1x_2\leq |f(t)-1|V$$
Using now Gronwall-Bellman inequality we obtain $$V(t)\leq V(0)\exp\left(\int_0^t|f(\tau)-1|d\tau\right)$$ from which stability follows if $$\int_0^\infty|f(\tau)-1|d\tau\leq C$$ for some $C>0$.