Here's an interesting problem. Consider a DC voltage source (for bias), an inductor (supply wires, for instance) and a resistor, which changes its resistance as a function of time: \$ R(t) = R_{offset}+R_{mod}\cos(\omega t)\$, where \$R_{offset} > R_{mod}\$. The bias voltage across the photoresistor would cause an AC-current \$i(t)\$ to flow.
simulate this circuit – Schematic created using CircuitLab
Now, to find the unknown AC-current \$i(t)\$, I would solve the following ODE: $$ V=L\frac{di}{dt}+i(t)R(t) $$ which can be rearranged to: $$ \frac{di}{dt}+i(t)\frac{R(t)}{L}=\frac{V}{L} $$ which enables us to solve the DE using the integration factor method. $$ i(t)=\frac{\int e^{\int p(t)dt}g(t)dt+c}{e^{\int p(t)dt}} $$ where \$g(t)=\frac{V}{L}\$ and \$p(t)=\frac{R_{offset}}{L}+\frac{R_{mod}}{L}\cos(\omega t)\$. We get:
$$ i(t)=\frac{\frac{V}{L}\int e^{\big(\frac{R_{offset}}{L}t+\frac{R_{mod}}{\omega L}\sin(\omega t)\big)}dt + c}{e^{\big(\frac{R_{offset}}{L}t+\frac{R_{mod}}{\omega L}\sin(\omega t)\big)}} $$ which becomes: $$ \require{cancel} i(t)=\frac{V}{\cancel{L}}\frac{\cancel{L}}{R_{offset}t + \frac{R_{mod}}{\omega}\sin(\omega t)} + \frac{c}{e^{\big(\frac{R_{offset}}{L}t+\frac{R_{mod}}{\omega L}\sin(\omega t)\big)}}\\ =\frac{V}{R_{offset}t + \frac{R_{mod}}{\omega}\sin(\omega t)} + \frac{c}{e^{\big(\frac{R_{offset}}{L}t+\frac{R_{mod}}{\omega L}\sin(\omega t)\big)}}\\ $$ The first term goes to 0 after briefly going to \$\infty \$ when \$R_{offset}t\$ and \$\frac{R_{mod}}{\omega L}\sin(\omega t)\$ are equal (big L-'hard-start'?). The other term goes to 0 with \$e^{-t}\$ which looks kind of strange, as I would expect some sort of steady-state solution.
Am I asking the wrong kind of question, here? i.e., is the problem badly stated, or am I missing something else?
EDIT: After doing the write-up, I have a suspicion, that: $$ \int e^{\big( \frac{R_{offset}}{L}t+\frac{R_{mod}}{\omega L}\sin(\omega t)\big)}dt \neq \frac{L}{R_{offset}t + \frac{R_{mod}}{\omega}\sin(\omega t)}e^{\big(\frac{R_{offset}}{L}t+\frac{R_{mod}}{\omega L}\sin(\omega t)\big)} $$ I will have another look, and correct if needed.