My solution to inhomogeneous $\frac{d^2y}{dx^2} + y = \sin{x}$ does not conform to my book's solution! I need help with the solution of this particular equation:

$$\frac{d^2y}{dx^2} + y = \sin{x}$$

Due to me having to go to work, I cannot display all my work in mathjax, my shift starts in 5 min...but my solution is: $$y= \frac{-x}{2} + \frac{(\cos{2x}\sin{x})}{4} + C_1\cos{x} + C_2\sin{x}$$
Whereas my book's solution is: $$y= \frac{-x}{2} +C_1\cos{x} + C_2\sin{x}$$
I have used the method of variation of prameters, cramer's rule and basic antidifferentiation to solve resulting system.
Thank you all!

Since you mention variation of parameters, I'll summarize the method here. Hopefully you can retrace your steps to find the mistake.
From the fundamental solution
$$ y_h(x) = c_1\cos x + c_2\sin x $$
we then seek a particular solution of the form
$$ y_p(x) = u_1(x)\cos x + u_2(x)\sin x $$
Plugging in the solution and following the method, we arrive at the system of equations:
\begin{align} {u_1}'\cos x + {u_2}'\sin x &= 0 \\ -{u_1}'\sin x + {u_2}'\cos x &= \sin x \end{align}
Using Kramer's rule, we find
\begin{align} {u_1}' &= -\sin^2 x = \frac{-1+\cos 2x}{2} \\ {u_2}' &= \cos x\sin x = \frac{\sin 2x}{2} \end{align} 
Integrating gives
\begin{align} y_p(x) &= \left(-\frac{x}{2} + \frac{\sin 2x}{4} \right)\cos x - \frac{\cos 2x}{4}\sin x \\
&= -\frac{x}{2}\cos x + \frac{\sin 2x\cos x-\cos 2x\sin 2x}{4} \\
&= -\frac{x}{2}\cos x +\frac{\sin x}{4} \end{align}
We can disregard the second term, since it'll combine with the fundamental solution. Thus, the general solution is
$$ y(x) = -\frac{x}{2}\cos x + c_1\cos x + c_2\sin x $$
