Question Number 107144 by mathmax by abdo last updated on 09/Aug/20
$$\mathrm{solve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{''} −\mathrm{xy}^{'} \:+\mathrm{2y}\:=\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right) \\ $$
Answered by bemath last updated on 09/Aug/20
$${y}={x}^{{r}} \:\rightarrow\begin{cases}{{y}'={rx}^{{r}−\mathrm{1}} }\\{{y}''={r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow{x}^{{r}} \left({r}^{\mathrm{2}} −{r}−{r}+\mathrm{2}\right)=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}+\mathrm{2}=\mathrm{0}\:;\:\left({r}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${r}=\mathrm{1}\pm{i}\:\Rightarrow{y}={e}^{{x}} \left({C}_{\mathrm{1}} \mathrm{cos}\:{x}+{C}_{\mathrm{2}} \mathrm{sin}\:{x}\right) \\ $$