Differential equation of $y'+y\sin(x) = \sin(2x) $ where $y=y(x)$?

Question

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

Answer 1

$$ y'+y\sin x=\sin (2x)\quad\Longleftrightarrow\quad \mathrm{e}^{-\cos x}(\,y'+y\sin x)=2 \sin x\cos x\,\mathrm{e}^{-\cos x} \\ \quad\Longleftrightarrow\quad \big(\mathrm{e}^{-\cos x}y\big)'=f(\cos x)\sin x=-\frac{d}{dx}F(\cos x), $$ 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 $c\in\mathbb R$.

Answer 2

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

Let $\quad X=\cos(x)$ $$-\frac{dy}{dX}+y=2X$$ This first order linear ODE is easy to solve : $\quad y=c\:e^X+2X+2$ $$y(x)=c\:e^{\cos(x)}+2\cos(x)+2$$