Solve $y''+4y=10\sin(2x)$ with an initial condition of $y(0)=y'(0)=0$. 
Solve $y''+4y=10\sin(2x)$ with an initial condition of $y(0)=y'(0)=0$.

Here is what I have done:

I could not make any sense after the last line. I need to find $A$ and $B$ from the equation and $C_1$ and $C_2$ and the solution equation.
 A: The guess for the  particular solution should be:
$$y_p=Ax \cos (2x)$$
Differentiate twice:
$$y_p''=-4A \sin (2x)-4Ax \cos (2x)$$
Plug this in the differential equation,  it gives:
$$y_p''+4y_p=-4A \sin (2x)$$
$$A=-\dfrac 5 2$$
$$\implies y_p=-\dfrac 52 x \cos (2x)$$
And:
$$\boxed {y(x)=c_1 \sin(2x) +c_2 \cos(2x) -\dfrac 52 x \cos (2x)}$$
A: You have $\sin(2x)$ and $\cos(2x)$ and NOT $\cos(x)$ and $\sin(x)$, so maybe that is where the confusion comes from.
So you have $$y'_{p}=A\cos(\color{red}{2x})-2Ax\sin(2x)+B\sin(\color{red}{2x})+2Bx\cos(2x)$$ and $$y''_{p}=-2A\sin(2x)-2A\sin(2x)-4Ax\cos(2x)+2B\cos(2x)+2B\cos(2x)-4Bx\sin(2x)$$
so substituting gives $$y''_{p}+4y_{p}$$
$$=-2A\sin(2x)-2A\sin(2x)-4Ax\cos(2x)+2B\cos(2x)+2B\cos(2x)-4Bx\sin(2x)$$
$$+4Ax\cos(2x)+4Bx\sin(2x)$$
$$=-4A\sin(2x)+4B\cos(2x)=10\sin(2x)$$
So $A=-\frac{10}{4}=-\frac{5}{2}$ and $B=0.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\mrm{y}''\pars{x} + 4\,\mrm{y}\pars{x} = 10\sin\pars{2x}.\quad \mrm{y}\pars{0}= \mrm{y}'\pars{0} = 0}:
\ {\Large ?}}$

Lets $\ds{\mrm{z}\pars{x} \equiv
\mrm{y}'\pars{x} + 2\,\mrm{y}\pars{x}\ic \implies
\bbx{\mrm{y}\pars{x} = {1 \over 2}\,\Im\mrm{z}\pars{x}}}$
$$
\implies\mrm{z}'\pars{x} - 2\ic\,\mrm{z}\pars{x} =
10\sin\pars{2x}\,,\qquad \mrm{z}\pars{0} = 0
$$
Then,
\begin{align}
\expo{-2\ic x}\mrm{z}'\pars{x} -
2\ic\expo{-2\ic x}\mrm{z}\pars{x} & =
10\expo{-2\ic x}\sin\pars{2x}
\\[5mm]
\totald{\bracks{\expo{-2\ic x}\mrm{z}\pars{x}}}{x} & =
10\expo{-2\ic x}\sin\pars{2x}
\\[5mm]
\expo{-2\ic x}\mrm{z}\pars{x} & =
10\int_{0}^{x}\expo{-2\,\ic\, t}\sin\pars{2t}\dd t
\\[5mm]
\expo{-2\ic x}\mrm{z}\pars{x} & =
{5 \over 4} - {5 \over 4}\,\expo{-4\ic x} - 5\ic x
\\[5mm]
\mrm{z}\pars{x} & =
{5 \over 4}\,\expo{2\ic x} - {5 \over 4}\,\expo{-2\ic x} - 5\ic x\expo{2\ic x}
\end{align}

$$
\mrm{y}\pars{x} = {5 \over 8}\,\sin\pars{2x} -
{5 \over 8}\bracks{-\sin\pars{2x}} -
{5 \over 2}\,x\cos\pars{2x}
$$
$$
\bbx{\mrm{y}\pars{x} = {5 \over 4}\,\sin\pars{2x} -
{5 \over 2}\,x\cos\pars{2x}} \\
$$
