Differential equation with particular solution ansatz terms same as homogenous solution

Question

so I am having a problem understanding finding solution to that equation: $$x'' +w^2 x =f sin ω t$$ with initial conditions x(0)=0 and x'(0)=0. I know that is an explanation of how to solve that:

I had to make the euler formula substitution and then multiply it by t . Ostensibly, this is because the the particular solution ansatz for fsintwt has a term(s) the same as the homogeneous solution. Hence Yp=Cte^(iwt). So, presumably, any g(t) ansatz that gives terms the same as the homogeneous solution must be adjusted accordingly? Was there some simpler way to solve this problem that I over looked? Please excuse any terminology errors in the preceding.

And

f≠0. This is the case of resonance, so you need to multiply your ansatz by t. So for the particular solution, you need to try $x=Atsinωt + Btcosωt$

But I do not understand what they are talking about. What I understand is that the general solution is: $$C_1 cos\omega t + C_2sin\omega t + x_(particular)$$ where to find x(particular) I need to make cost = x, but then I come to f = 0 when I solve the left side of the equation. What am I doing wrong, not understanding, please help:) I really appreciate help as I spent a good few hours on that one problem, and very very thanks in advance!

Answer 1

What they mean is that when $\omega\neq 0$ and $f\neq 0$, the equation $x'' + \omega^2 x = f \sin (\omega t)$ has a particular solution of the form \begin{equation} x(t) = A t \sin(\omega t) + B t \cos(\omega t) \end{equation} You need to find $A$ and $B$ by substituting this solution in the equation.

Answer 2

A solution of the homogeneous equation will never work as a solution of the non-homogeneous equation for a very simple reason: when plugged in the LHS, it yields $0$ !

So when the RHS belongs to the homogeneous equation, the ansatz must be different. In the case of an equation with constant coefficients, the ansatz will be a polynomial of appropriate degree times the homogeneous solution.

Here we can try $$(C_1t+C_0)\cos\omega t+(S_1t+S_0)\sin\omega t,$$ which gives

$$2\omega S_1\cos(\omega t)-2\omega C_1\sin(\omega t)=f\sin(\omega t).$$