Second order homogeneous variable coefficient ODE (periodic perturbation) I have trouble solving the following ODE: $$y''(t)+[\alpha -i\beta \cos(\omega t)]y'(t)-\alpha^2 y(t)=0.$$ 
where $$\alpha\gg\beta, \alpha\gg\omega$$
I was wondering is there any approach to get an approximate solution to this ODE. 
I have tried the exploratory solution like this：$$y(x)=Y(x)*e^{-i\frac{\beta}{\omega}\sin\omega t}$$, but it's not work.
 A: You can apply techniques of perturbation theory as follows:
Let $y$ be written in terms of a Taylor series of a small parameter, which in your case is appropriately $\beta$. So, $y  = Y_0 + \beta Y_1 + \beta^2Y_2 + O(\beta^3)$, where $Y_i$ are functions of $t$ yet to be determined.
Then this substituting this into the original equation, $y'' + (\alpha - i\beta cos(\omega t))y' - \alpha^2y = 0$, we get the following
$$(Y_0’’+\alpha Y_0’- \alpha^2 Y_0) + \beta(Y_1’’+\alpha Y_1’-\alpha^2Y_1 - icos(\omega t)Y_0)+\beta^2(Y_2’’+\alpha Y_2’- \alpha^2 Y_2 - icos(\omega t)Y_1) + O(\beta^3) = 0$$
Equating each coefficient of the powers of $\beta$ to 0, results in a system of equations. Notably:
$$Y_0''+\alpha Y_0' - \alpha^2 Y_0=0$$
$$Y_1''+\alpha Y_1' - \alpha^2Y_1 - icos(\omega t)Y_0 =0$$
$$Y_2''+\alpha Y_2' - \alpha^2 Y_2 - icos(\omega t)Y_1 = 0$$
Now, the solution to $Y_0$ is $Y_0 = c_1e^{\gamma_1t}+c_2e^{\gamma_2t}$, where $\gamma_i$ are the roots of the equation $\gamma^2+\alpha \gamma - \alpha^2=0$.
You can then solve for $Y_1$ and $Y_2$ successively. I am not going to do all of the calculation work, but suffice it to say, there is a lot of algebra. For instance, take $Y_1$:
$$Y_1''+\alpha Y_1' - \alpha^2 Y_1 = icos(\omega t) Y_0= {i\over2}(e^{-i\omega t}+e^{i \omega t})(c_1e^{\gamma_1t}+c_2e^{\gamma_2t}).$$ There are four unique exponential terms on the right hand side. Consider one of them, $c_1e^{(i\omega+\gamma_1)t}$, and let $Y_{11}=d_1e^{(i\omega+\gamma_1)t}$. Then $d_1 = {i\over2}\frac{c_1}{-\alpha^2+\alpha(i\omega+\gamma_1)+(i\omega+\gamma_1)^2}={i\over2}\frac{c_1}{-\omega^2+i(\alpha + 2\gamma_1)\omega}$. 
In this fashion, you can calculate $Y_1 = \sum_{i=1}^4 Y_{1i}$.
After you have solved the $Y_i$, then $y \approx Y_0+\beta Y_1 +\beta^2 Y_2$.
