Question Number 46609 by maxmathsup by imad last updated on 29/Oct/18
$${solve}\:\:\:\:{x}\:{y}^{''} \:−{e}^{−{x}} {y}^{'} \:\:\:={x}\:{sinx} \\ $$
Commented by maxmathsup by imad last updated on 11/Nov/18
$${hangement}\:{y}^{'} ={z}\:{give}\:{xz}^{'} −{e}^{−{x}} {z}\:={xsinx} \\ $$$$\left({he}\right)\Rightarrow{xz}^{'} \:−{e}^{−{x}} {z}\:=\mathrm{0}\:\Rightarrow{xz}^{'} \:={e}^{−{x}} {z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{{e}^{−{x}} }{{x}}\:\Rightarrow{ln}\mid{z}\mid=\int\:\frac{{e}^{−{x}} }{{x}}{dx}\:+\alpha\:\Rightarrow \\ $$$${z}\:={K}\:{e}^{\int\frac{{e}^{−{x}} }{{x}}{dx}} \:\:{mvc}\:{method}\:{gve}\:{z}^{'} ={K}^{'} \:{e}^{\int\:\frac{{e}^{−{x}} }{{x}}{dx}} \:+{K}\:\frac{{e}^{−{x}} }{{x}}\:{e}^{\int\:\frac{{e}^{−{x}} }{{x}}{dx}} \\ $$$$\left({e}\right)\:\Rightarrow{xK}^{'} \:{e}^{\int\:\frac{{e}^{−{x}} }{{x}}{dx}} \:\:+{Ke}^{−{x}} \:{e}^{\int\:\frac{{e}^{−{x}} }{{x}}{dx}} \:−{e}^{−{x}} {K}\:{e}^{\int\frac{{e}^{−{x}} }{{x}}{dx}} \:={xsinx}\:\Rightarrow \\ $$$${xK}^{'} \:{e}^{\int\frac{{e}^{−{x}} }{{x}}{dx}} \:={xsinx}\:\Rightarrow{K}^{'} \:\:={sinx}\:{e}^{−\int\:\frac{{e}^{−{x}} }{{x}}{dx}} \:\Rightarrow\: \\ $$$${K}\left({x}\right)=\:\int_{.} ^{{x}} \:\left({sint}\:{e}^{−\int\:\frac{{e}^{−{t}} }{{t}}{dt}} \right){dt}\:+\lambda\:\Rightarrow{z}\:=\left(\int_{.} ^{{x}} \left({sint}\:{e}^{−\int\frac{{e}^{−{t}} }{{t}}{dt}} \right){dt}\:+\lambda\right){e}^{\int\:\frac{{e}^{−{x}} }{{x}}} \:{dx} \\ $$$${but}\:{y}^{'} \left({x}\right)={z}\left({x}\right)\:\Rightarrow{y}\left({x}\right)=\int\:{z}\left({x}\right){dx}\:+\beta\:\:{and}\:{the}\:{function}\:{z}\:{is}\:{determined}. \\ $$