How is Mathematica solving this ODE with periodic coefficients?

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Homework Statement
##y''(x)=a\cdot \cos{(\omega x)}(b+c\cdot y(x))##
Relevant Equations
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245639

Mathematica gives this solution but how does it calculate it? What's the method here?
 
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If there is a particular method or if this is a specific type of ode that lends itself to a straight forward solution, then I don't off hand, know what it is. Sooo, what I would generally do after rearranging it and trying to change variables into something I recognize is to just find a series solution and see if the recursion relation or the resulting terms can be collected into some standard functions. Personally, I think I'd go for leaving it as a series given the output from mathematica, which doesn't seem very illuminating.
 
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bobob said:
If there is a particular method or if this is a specific type of ode that lends itself to a straight forward solution, then I don't off hand, know what it is. Sooo, what I would generally do after rearranging it and trying to change variables into something I recognize is to just find a series solution and see if the recursion relation or the resulting terms can be collected into some standard functions. Personally, I think I'd go for leaving it as a series given the output from mathematica, which doesn't seem very illuminating.
Képkivágás.PNG

I wasn't able to derive a recursion relation due to the term ##y(x) \cdot \cos{(\omega x)}## as in power series form I get an infinite sum times an infinite sum, not sure how to deal with that. Calculating the coefficients one by one didn't look like a useful thing to do, as Mathematica can do it too. Moreover, the power series doesn't seem to converge even in higher orders (I tried it for 30), and it does not match the numerical solution:
Képkivágás.PNG

Does this mean that this cannot be solved with power series or I missed something?
 
I am not an expert on this, but ODEs with periodic coefficients are handled with Floquet theory. Your particular equation is essentially Mathieu’s equation
https://en.m.wikipedia.org/wiki/Mathieu_functionThe homogeneous equation has two solutions (of course), which are called Matthieu functions.

You have a non-homogeneous version ( your sinusoid forcing function). So Mathematica’s solution looks like what you would get if you applied variation of parameters using the two solutions to the homogeneous equation.

Jason