For ordinary differential equation, 

$$y'+y\sin(x) = \sin(2x) $$

where y=y(x) for real x. 

Is there any way that I can solve this question with eigenvalues and eigenvectors by changing above equation like X`=AX ? 

Thanks in advance
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$$\frac{dy}{dx}+y\sin(x) = \sin(2x)=2\sin(x)\cos(x) $$
$$\frac{dy}{\sin(x)dx}+y =2\cos(x)$$
$$\frac{dy}{d(-\cos(x))}+y=2\cos(x) $$

Let $\quad X=\cos(x)$
$$-\frac{dy}{dX}+y=2X$$
This first order linear ODE is easy to solve : $\quad y=c\:e^X+2X+2$
$$y(x)=c\:e^{\cos(x)}+2\cos(x)+2$$
$$
y'+y\sin x=\sin (2x)\quad\Longleftrightarrow\quad
\mathrm{e}^{-\cos x}(\,y'+y\sin x)=2 \sin x\cos x\,\mathrm{e}^{-\cos x}
\\ \quad\Longleftrightarrow\quad
\big(\mathrm{e}^{-\cos x}y\big)'=f(\cos x)\sin x=-\frac{d}{dx}F(\cos x),
$$
where
$$
f(w)=2 w\mathrm{e}^{-w}\quad\text{and}\quad F(w)=\int f(w)\,dw.
$$
But
$$
F(w)=\int 2 w\,\mathrm{e}^{-w}\,dw=-2(w+1)\mathrm{e}^{-w}+c
$$
and hence
$$
\big(\mathrm{e}^{-\cos x}y\big)'=-\frac{d}{dx}\left(-2(\cos x+1)\mathrm{e}^{-\cos x}\right)
$$
and finally
$$
\mathrm{e}^{-\cos x}y=2(\cos x+1)\mathrm{e}^{-\cos x}+c
$$
or
$$
y=2(\cos x+1)+c\,\mathrm{e}^{\cos x}
$$
for some $c\in\mathbb R$.