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Question Number 90973 by abdomathmax last updated on 27/Apr/20 | ||
$${solve}\:{xy}^{'} \:+\left({x}+\mathrm{1}\right){y}\:={e}^{−{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$ | ||
Commented by mathmax by abdo last updated on 27/Apr/20 | ||
$$\left({he}\right)\rightarrow{xy}^{'} \:+\left({x}+\mathrm{1}\right){y}\:=\mathrm{0}\:\Rightarrow{xy}^{'} \:=−\left({x}+\mathrm{1}\right){y}\:\Rightarrow\frac{{y}^{'} }{{y}}=−\frac{{x}+\mathrm{1}}{{x}} \\ $$$$=−\mathrm{1}−\frac{\mathrm{1}}{{x}}\:\Rightarrow{ln}\mid{y}\mid\:=−{x}−{ln}\mid{x}\mid\:+{c}\:\Rightarrow{y}\left({x}\right)={k}\:{e}^{−{x}} ×\frac{\mathrm{1}}{\mid{x}\mid} \\ $$$$\left.{solution}\:{on}\:\right]\mathrm{0},+\infty\left[\:\Rightarrow{y}\left({x}\right)\:={k}\:\frac{{e}^{−{x}} }{{x}}\:\:{let}\:{use}\:{mvc}\:{method}\right. \\ $$$${y}^{'} \:={k}^{'} \:\frac{{e}^{−{x}} }{{x}}\:+{k}\left\{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{e}^{−{x}} \:−\frac{\mathrm{1}}{{x}}{e}^{−{x}} \right\} \\ $$$$\left({e}\right)\Rightarrow{k}^{'} \:{e}^{−{x}} \:−\frac{{k}}{{x}}{e}^{−{x}} −{ke}^{−{x}} \:+{k}\left({x}+\mathrm{1}\right)\frac{{e}^{−{x}} }{{x}}\:={e}^{−{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${k}^{'} \:{e}^{−{x}} \:−\frac{{k}}{{x}}{e}^{−{x}} \:−{k}\:{e}^{−{x}} \:+{ke}^{−{x}} \:+\frac{{k}}{{x}}{e}^{−{x}} \:={e}^{−{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${k}^{'} \:={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:\Rightarrow{k}\left({x}\right)\:=\int{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}\:+{c} \\ $$$${but}\:\int\:{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}\:=_{{byparts}} \:\:\:{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\int{x}\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{2}\:\int\:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{2}+\mathrm{2}{arctan}\left({x}\right)\:\Rightarrow \\ $$$${k}\left({x}\right)={xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\mathrm{2}{arctan}\left({x}\right)\:+{C}\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\frac{{e}^{−{x}} }{{x}}{k}\left({x}\right)\:=\frac{{e}^{−{x}} }{{x}}\left({xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\mathrm{2}{arctan}\left({x}\right)\:+{C}\right) \\ $$ | ||
Commented by mathmax by abdo last updated on 27/Apr/20 | ||
$${forgive}\:\int\:{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}\:={xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}\:+\mathrm{2}{arctan}\left({x}\right)\:+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\frac{{e}^{−{x}} }{{x}}\left(\:{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}\:+\mathrm{2}{arctan}\left({x}\right)+{C}\right) \\ $$ | ||
Answered by Joel578 last updated on 27/Apr/20 | ||
$${y}'\left({x}\right)\:+\:{p}\left({x}\right){y}\left({x}\right)\:=\:{r}\left({x}\right) \\ $$$$\mathrm{Has}\:\mathrm{the}\:\mathrm{solution} \\ $$$${y}\left({x}\right)\:=\:\frac{\mathrm{1}}{{F}\left({x}\right)}\left[\int\:{F}\left({x}\right){r}\left({x}\right)\:{dx}\:+\:{C}\right] \\ $$$$\mathrm{where}\:{F}\left({x}\right)\:\mathrm{is}\:\mathrm{integrating}\:\mathrm{factor}\:\mathrm{and}\:{C}\:\mathrm{is}\:\mathrm{integration}\:\mathrm{constant} \\ $$$$ \\ $$$${y}'\:+\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\right){y}\:=\:\frac{{e}^{−{x}} \:\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}} \\ $$$$\mathrm{Integrating}\:\mathrm{factor}\: \\ $$$${F}\left({x}\right)\:=\:\mathrm{exp}\left[\int\:\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\:{dx}\right]\:=\:\mathrm{exp}\left({x}\:+\:\mathrm{ln}\:{x}\right)\:=\:{xe}^{{x}} \\ $$$$\mathrm{hence} \\ $$$${y}\left({x}\right)\:=\:\frac{{e}^{−{x}} }{{x}}\left[\int\:{xe}^{{x}} .\frac{{e}^{−{x}} \:\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}\:{dx}\:+\:{C}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{e}^{−{x}} }{{x}}\left[\int\:\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:{dx}\:+\:{C}\right] \\ $$$$ \\ $$$$\mathrm{Use}\:\mathrm{integration}\:\mathrm{by}\:\mathrm{part}:\:{u}\:=\:\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right),\:{dv}\:=\:{dx}\: \\ $$$${I}\:=\:{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:−\:\int\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:=\:{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:+\:\int\:\frac{\mathrm{2}\:−\:\mathrm{2}\:−\:\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$$$\:\:\:=\:{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:+\:\int\:\frac{\mathrm{2}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:−\:\mathrm{2}\:{dx} \\ $$$$\:\:\:=\:{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:+\:\mathrm{2tan}^{−\mathrm{1}} {x}\:−\:\mathrm{2}{x} \\ $$$$ \\ $$$$\therefore\:{y}\left({x}\right)\:=\:\frac{{e}^{−{x}} }{{x}}\left[{x}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:+\:\mathrm{2tan}^{−\mathrm{1}} {x}\:−\:\mathrm{2}{x}\:+\:{C}\right] \\ $$ | ||
Commented by mathmax by abdo last updated on 27/Apr/20 | ||
$${thank}\:{you}\:{sir}. \\ $$ | ||