Question Number 78999 by gopikrishnan last updated on 22/Jan/20
$${if}\:{dy}/{dx}+{y}=\mathrm{1}+{y}/{x},{then}\:{the}\:{integrating}\:{factor}\:\left({I},{F}\right)\:{is} \\ $$
Commented by mr W last updated on 22/Jan/20
$${please}\:{don}'{t}\:{type}\:{the}\:{whole}\:{post}\:{in}\:{a} \\ $$$${single}\:{line}\:{sir}! \\ $$$$ \\ $$$${with}\:\mathrm{1}+{y}/{x}\:{do}\:{you}\:{mean}\:\mathrm{1}+\frac{{y}}{{x}}\:{or} \\ $$$$\frac{\mathrm{1}+{y}}{{x}}? \\ $$
Commented by jagoll last updated on 22/Jan/20
$$\mathrm{not}\:\mathrm{clear}\:\mathrm{sir}\:\mathrm{your}\:\mathrm{question} \\ $$
Answered by mind is power last updated on 22/Jan/20
$${if}\:\frac{{dy}}{{dx}}+{y}=\mathrm{1}+\frac{{y}}{{x}},{x}\in\mathbb{R}^{\ast} \\ $$$$\Leftrightarrow\frac{{dy}}{{dx}}+{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}…{W} \\ $$$${Homgenius} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}+{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\frac{{dy}}{{y}}=\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){dx},{x} \\ $$$$\Rightarrow{ln}\left({y}\right)=\left({ln}\left({x}\right)−{x}\right) \\ $$$$\Rightarrow{y}={kxe}^{−{x}} ,{k}\:{Constante} \\ $$$${General}\:{y}_{{p}} \left({x}\right)={k}\left({x}\right){xe}^{−{x}} ,{Solution}\:{of}\:..{W} \\ $$$${k}'{xe}^{−{x}} +\left({e}^{−{x}} −{xe}^{−{x}} \right){k}\left({x}\right)+{kxe}^{−{x}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{1} \\ $$$$\Rightarrow{k}'\left({x}\right)=\frac{{e}^{−{x}} }{{x}}\Rightarrow{k}\left({x}\right)=\int\frac{{e}^{−{x}} }{{x}}=−{E}_{{i}} \left({x}\right),\:{Exponentiel}\:{Integral} \\ $$$${Y}_{{p}} =\left(−{E}_{{i}} \left({x}\right)\right){xe}^{−{x}} \\ $$$${Y}={Y}_{{h}} +{Y}_{{p}} ={kxe}^{−{x}} −{E}_{{i}} \left({x}\right){xe}^{−{x}} ,{k}\in\mathbb{R},{x}\in\mathbb{R}_{+} ^{\ast} \\ $$$$\frac{{dy}}{{dx}}+{y}=\frac{\mathrm{1}+{y}}{{x}} \\ $$$$\Leftrightarrow{x}\frac{{dy}}{{dx}}={y}\left(\mathrm{1}−{x}\right)+\mathrm{1} \\ $$$${Homgenius} \\ $$$$\frac{{dy}}{{dx}}.\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}−{x}}{{x}}\Rightarrow{ln}\left({y}\right)=\left({ln}\left({x}\right)−{x}\right)+{c}\Rightarrow{y}={kxe}^{−{x}} \\ $$$${let}\:{Y}_{{p}} ={k}\left({x}\right){xe}^{−{x}} \:{particular}\:{Solution}\Rightarrow \\ $$$${x}\left({k}'{xe}^{−{x}} +{k}\left({e}^{−{x}} −{x}\right)\right)+\left({x}−\mathrm{1}\right){kxe}^{−{x}} =\mathrm{1} \\ $$$$\Rightarrow{k}'=\frac{{e}^{−{x}} }{{x}^{\mathrm{2}} }\Rightarrow{k}=\int\frac{{e}^{−{x}} }{{x}^{\mathrm{2}} },{by}\:{Part}=\left(−\frac{{e}^{−{x}} }{{x}}−\int\frac{{e}^{−{x}} }{{x}}{dx}\right. \\ $$$$=−\frac{{e}^{−{x}} }{{x}}−{E}_{{i}} \left({x}\right) \\ $$$${Y}_{{p}} =\left(−\frac{{e}^{−{x}} }{{x}}−{E}_{{i}} \left({x}\right)\right){xe}^{−{x}} +{kxe}^{−{x}} ,{k}\in\mathbb{R} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$