Question Number 128951 by mathmax by abdo last updated on 11/Jan/21
$$\mathrm{solve}\:\mathrm{y}^{,,} −\mathrm{2y}^{'} \:+\mathrm{3y}\:=\mathrm{xe}^{−\mathrm{x}} \mathrm{sin}\left(\mathrm{2x}\right)\:\:\mathrm{with}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1} \\ $$
Answered by mnjuly1970 last updated on 11/Jan/21
![m^2 −2m+3=0 (characteristic equation) roots:: m_(1,2) =1+_− i y_(p ) =e^(−x) [(Ax+B)sin(2x)+(Cx+D)cos(2x)]](https://www.tinkutara.com/question/Q128960.png)
$$\:\:\:{m}^{\mathrm{2}} −\mathrm{2}{m}+\mathrm{3}=\mathrm{0}\:\left({characteristic}\:{equation}\right) \\ $$$$\:\:{roots}::\:\:{m}_{\mathrm{1},\mathrm{2}} =\mathrm{1}\underset{−} {+}\:{i} \\ $$$$\:\:{y}_{{p}\:} ={e}^{−{x}} \left[\left({Ax}+{B}\right){sin}\left(\mathrm{2}{x}\right)+\left({Cx}+{D}\right){cos}\left(\mathrm{2}{x}\right)\right] \\ $$$$ \\ $$