So I'm solving some practice questions and I have a differential equation $$y''-3y'+2y=\sin(x)$$ about $x=0$, and I'm told to solve it using the Frobenius method. I don't think that's possible because $x=0$ would be an ordinary point rather than a singular point, so I'm solving it by using a series solution method. However, I think I'm forgetting how you use it for a nonhomogenous equation? What do I do with the $sin(x)$ that I have at the RHS?
My current progress is below:
$$y(x) = \sum_{n=0}^{\infty} a_nx^n $$
$$y'(x) = \sum_{n=1}^{\infty} na_nx^{n-1} $$
$$y''(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}$$
Then, putting it into the equation and then simplifying, I get:
$$\sum_{n=0}^{\infty}\bigg[(n+2)(n+1)a_{n+2} - 3(n+1)a_{n+1} + 2a_n \bigg]x^n = \sin(x)$$
I have two questions about this. Should I solve it homogenously first and get an expression for the $a_n$'s and stuff, or should I keep the $\sin(x)$ there? And if I do, what do I do with it? Like how do I get a proper expression that'll give me a solution to the DE? I've been trying to find examples online but most of them only have the RHS equal to constants or powers of x, and no other functions, which is why I'm kind of confused. Any help at all would make me super super grateful!!
HINT
Here is another method you can compare with: \begin{align*} y'' - 3y' + 2y = \sin(x) & \Longleftrightarrow (y' - 2y)' - (y' - 2y) = \sin(x)\\\\ & \Longleftrightarrow u' - u = \sin(x)\\\\ & \Longleftrightarrow [\exp(-x)u]' = \exp(-x)\sin(x)\\\\ & \Longleftrightarrow \exp(-x)u = -\exp(-x)\left[\frac{\cos(x) + \sin(x)}{2}\right] + k\\\\ & \Longleftrightarrow y' - 2y = -\frac{\cos(x) + \sin(x)}{2} + k\exp(x)\\\\ & \Longleftrightarrow [\exp(-2x)y]' = -\exp(-2x)\left[\frac{\cos(x) + \sin(x)}{2}\right] + k\exp(-x)\\\\ \end{align*}
Can you take it from here?
EDIT
The power series of the sine function is given by: \begin{align*} \sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots \end{align*}