DSolve[
  {x''[t] + (k/2 - i(d + g(c + a Cos[t ω]))) x'[t] + i g a ω Sin[t ω] x[t] == 0}, 
  x[t], t]

I tried with initial condition x[0] == x'[0] == 0 just to test it, but it's showing an error.

The solution is in terms of Fourier series with coefficients having Bessel functions of the first kind. $Edit$: Initially, its a first order differential equation.

x'[t]=-k/2(x[t]-x1)+i(d +g(c+ a Cos[t ω])) x[t], 

where d, x1, c, g and a are constant. Since, the solution shows some Bessel function of first kind. So, I thought it requires a 2nd order differential equation to get that solution. But not sure if it is possible with first order derivative equation as well. so I differentiate the equation to get double derivative w.r.t. time as shown inside the DSolve. Its actual solution is written in a Fourier series: $$x(t)=e^{\text{i$\phi $} (t)} \sum _n x_n e^{i n \omega t}$$ where coefficient is $$x_{n}=\frac{x1}{2} \frac{J_{n}(-g a/\omega)}{i n \omega/k +{1/2} -i(g c+ d)/k}$$ and $$\phi = (g a/\omega)Sin(\omega t)$$.

closed as off-topic by AccidentalFourierTransform, Lukas Lang, Mariusz Iwaniuk, Bob Hanlon, Henrik Schumacher Nov 10 at 17:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – AccidentalFourierTransform, Lukas Lang, Mariusz Iwaniuk, Bob Hanlon, Henrik Schumacher
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  • 3
    Cos instead of cos and \[Omega] instead of \omega. – AccidentalFourierTransform Nov 10 at 16:46
  • 1
    With the corrections made to the code, DSolve provides an answer. – bbgodfrey Nov 11 at 10:26
  • @bbgodfrey Any suggestion to get the solution in Fourier series? – Off Topic Nov 13 at 16:48
  • What boundary conditions do you wish? – bbgodfrey Nov 13 at 18:59
  • @bbgodfrey, I have added more detail but not sure about the boundary conditions. – Off Topic Nov 14 at 10:23