Consider the classical harmonic oscillator: $$ \ddot{q}+\omega^2q=0, $$ which has a solution $$ q(t)=a \cos (\omega t+\phi). $$ Let me take $\phi=0$ for simplicity. Suppose now that $\omega$, instead of being a constant, is allowed to vary slowly with $t$, and that this functional relation is known. The fact that $\omega$ can change slowly with $t$ is modeled by introducing a slow time variable $T=\epsilon t$, with $\epsilon$ small.
The new equation of motion is: $$ \ddot{q}+\omega^2(T)q=0. $$ Consider the ansatz: $$ q(t,T) = a(T) \cos(\omega(T)t). $$ Clearly, $$ \dot{q} = -a\omega\sin\omega t + \epsilon\left(a'\cos\omega t-a\omega't\sin\omega t\right), $$ and $$ \ddot{q} = -a\omega^2\cos\omega t - 2\epsilon\left(a\omega'\sin\omega t+a'\omega\sin\omega t + a\omega\omega't\cos\omega t\right) + \epsilon^2\left(a''\cos\omega t - 2a'\omega'\sin\omega t-a\omega'^2t^2\cos\omega t-a\omega'' t\sin\omega t\right), $$ and replacing in the equation of motion gives: $$ -2\epsilon\left(a\omega'\sin\omega t+ a'\omega\sin\omega t+ a\omega\omega' t\cos\omega t\right) + \epsilon^2\left(a''\cos\omega t-2a'\omega't\sin\omega t-a\omega'^2t^2\cos\omega t - a\omega'' t\sin\omega t\right) = 0. $$ Note that the term $O(1)$ is identically zero. I suppose that an equation for $a$ could be seeked by imposing that the term $O(\epsilon)$ vanishes as well. This would give: $$ \frac{a'}{a} = -\frac{\omega'}{\omega}-\omega'\cot\omega t, $$ which could be fine as $\omega$ is known, however the thing that puzzles me is that $a$ is a function of $T$, while on the right hand side there is a term which is a function of $t$. How to deal with this? Maybe my ansatz was wrong?