Question Number 67014 by mathmax by abdo last updated on 21/Aug/19
$${solve}\:{y}^{''} +{x}^{\mathrm{2}} {y}^{'} ={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right) \\ $$
Commented by mathmax by abdo last updated on 22/Aug/19
$${we}\:{use}\:{the}\:{changement}\:{y}^{'} ={z}\:\:\:\:\:\:\left({e}\right)\rightarrow{z}^{'} \:+{x}^{\mathrm{2}} {z}\:={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right) \\ $$$$\left({he}\right)\rightarrow{z}^{'} \:+{x}^{\mathrm{2}} {z}\:=\mathrm{0}\:\Rightarrow{z}^{'} =−{x}^{\mathrm{2}} {z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=−{x}^{\mathrm{2}} \:\Rightarrow{ln}\mid{z}\mid=−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+{c}\:\Rightarrow \\ $$$${z}\:=\:{K}\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:\:{let}\:{use}\:{mvc}\:{method}\:\:{z}^{'} ={K}^{'} \:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:−{x}^{\mathrm{2}} \:{K}\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \\ $$$$\left({e}\right)\:\Rightarrow\left({K}^{'} −{x}^{\mathrm{2}} {K}\:\right){e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:+{x}^{\mathrm{2}} {K}\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right)\:\Rightarrow \\ $$$${K}^{'} \:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:={e}^{−{x}} {sin}\left(\mathrm{3}{x}\right)\:\Rightarrow{K}^{'} \:={e}^{−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:{sin}\left(\mathrm{3}{x}\right)\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int\:\:\:{e}^{−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} {sin}\left(\mathrm{3}{x}\right){dx}+\lambda\:\:{by}\:{parts}\:\:\:{u}\:={e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{and}\:{v}={sin}\left(\mathrm{3}{x}\right) \\ $$$${K}\left({x}\right)\:={e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{sin}\left(\mathrm{3}{x}\right)−\mathrm{3}\int\:\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{cos}\left(\mathrm{3}{x}\right){dx} \\ $$$${by}\:{parts}\:\int\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} {cos}\left(\mathrm{3}{x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{2}} −\mathrm{1}\right){e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} {sin}\left(\mathrm{3}{x}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\:\left\{\:\mathrm{2}{x}\:{e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:\:+\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} {e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{1}} \right\}{sin}\left(\mathrm{3}{x}\right){dx}\:=…..{any}\:{way}\:{let} \\ $$$${take}\:{K}\left({x}\right)=\int\:\:{e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} \:{sin}\left(\mathrm{3}{x}\right){dx}\:+\lambda\Rightarrow{z}\left({x}\right)\:={e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \left(\int\:{e}^{\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−{x}} {sin}\left(\mathrm{3}{x}\right){dx}+\lambda\right) \\ $$$$=\lambda\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:+\:{e}^{−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}} \:\int^{{x}} \:{e}^{\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}} \:{sin}\left(\mathrm{3}{t}\right){dt} \\ $$$${y}^{'} ={z}\:\Rightarrow{y}\left({x}\right)\:=\int^{{x}} {z}\left({u}\right){du}\:+{c}\:=\int^{{x}} \left\{\:\lambda\:{e}^{−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}} \:+{e}^{−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}} \:\int^{{u}} \:{e}^{\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}} \:{sin}\left(\mathrm{3}{t}\right){dt}\right\}{du}+{c} \\ $$