Solve second order differential equation of resonance. 
Solve equation 
  $$ y'' + y = \sin x $$

$$y=e^{rx}$$
$$r^2e^{rx}+e^{rx}=0$$
$$r^2 +1 =0$$
$$r=-i, \ r =i \tag{1}$$
I use the equation 
$$y=y_0 + A e^{\alpha x} \cos\beta x + Be^{\alpha x} \sin\beta x$$ 
where from $(1)$: $\alpha = 0 ,\ \beta = 1 $ or  $\alpha = 0 ,\ \beta = -1 $
I get 
$$y = A \cos x + B \sin x $$ $$y = A \cos x - B \sin x$$ 
because $A,\ B$ are real constants the only solution is:
$$ y = y_0 + A \cos x + B \sin x ,\  A,B \in \mathbb{R} $$
$$ y_0 = u_1(x) \cos x + u_2(x) \sin x $$
$$ y_0' = u_1'\cos x -u_1\sin x + u_2' \sin x + u_2 \cos x \tag{2} $$
There is assumption which I don't understand: 
$$ u_1' \cos x + u_2'\sin x =0 $$
and with this assumption from $(2)$
$$ y_0'= -u_1\sin x + u_2\cos x $$ 
$$y_0'' = -u_1'\sin x -u_1\cos x + u_2'\cos x -u_2\sin x $$
And then we have two equations which we have to solve:
$$ u_1'\cos x +u_2'\sin x = 0 $$
$$ -u_1'\sin x +u_2'\cos x = \sin x $$
...
 A: $$y''+y=\sin(x)$$
This is a second order linear non-homogeneous ODE.
First, solve the associated second order linear homogeneous ODE :
$$Y''+Y=0 \quad\to\quad Y=A\cos(x)+B\sin(x)$$
$A$ and $B$ are constants.
Second, find a solution of the non-homogeneous ODE thanks to the variation of parameters method (for example). That is, remplace the constants by functions :
$$y_0(x)=f(x)\cos(x)+g(x)\sin(x)$$
Note that we don't need to solve for the general solution $y(x)$ but only for any one solution $y_0(x)$
A short way should be to guess a simple form for the functions $f(x)$ and $g(x)$, linear in the present case. Supposing that we cannot guess, then :
$y_0'=(f'+g)\cos(x)+(g'-f)\sin(x)$
$y_0''=(f''+2g'-f)\cos(x)+(g''-2f'-g)\sin(x)$
$y_0''+y_0=\sin(x)=(f''+2g'-f)\cos(x)+(g''-2f'-g)\sin(x)+f\cos(x)+g\sin(x)$
$(f''+2g')\cos(x)+(g''-2f')\sin(x)=\sin(x)\quad\to\quad 
\begin{cases}f''+2g'=0\\g''-2f'=1\end{cases}$
Of course, we could continue and use a sledgehammer to crack a nut! But since we don't need all solutions, but only one, obviously $\quad f=-\frac{x}{2}\quad $ and $\quad g=0\quad$ is sufficient.
$$y_0(x)=-\frac{x}{2}\cos(x)$$
The general solution of $y''+y=\sin(x)$ is : $\quad y(x)=y_0(x)+Y(x)$
$$y(x)=-\frac{x}{2}\cos(x)+A\cos(x)+B\sin(x)$$
