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!

$$y'' + y = \sin{x}$$
$$y''\cos(x) + y\cos(x) = \frac 12 \sin{(2x)}$$
$$y''\cos(x) -y'\sin(x)+y'\sin(x)+ y\cos(x) = \frac 12 \sin{(2x)}$$
$$(y'\cos(x))'+(y\sin(x))'= \frac 12 \sin{(2x)}$$
First integration
$$(y'\cos(x))+(y\sin(x))= -\frac 14 \cos{(2x)}+K_1$$
$$(\frac y {\cos(x)})'= -\frac 14 \frac {\cos{(2x)}+K_1}{\cos^2(x)}$$
Second integration
$$(\frac y {\cos(x)})= -\frac 14 \int  \frac {\cos{(2x)}+K_1}{\cos^2(x)}dx+K_2$$
$$(\frac y {\cos(x)})= -\frac 14 \int  \frac {2\cos^2(x)+K_1}{\cos^2(x)}dx+K_2$$
$$ y = -\frac 12x\cos(x)+ K_1\cos(x)\int  \frac {dx}{\cos^2(x)}+K_2\cos(x)$$
Finally
$$ \boxed{y = -\frac 12x\cos(x)+ K_1\sin(x)+K_2\cos(x)}$$
