Question Number 73488 by abdomathmax last updated on 13/Nov/19
$${solve}\:\:\:{xy}^{''} \:\:+\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{'} \:\:={x}\:{e}^{−{x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 14/Nov/19
$${let}\:\:{y}^{'} ={z}\:\:\:\left({e}\right)\:\Rightarrow{xz}^{'} \:+\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}\:={xe}^{−{x}^{\mathrm{2}} } \\ $$$$\left({he}\right)\rightarrow\:{xz}^{'} \:+\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}\:={o}\:\Rightarrow{xz}^{'} \:=\left(\mathrm{1}−{x}^{\mathrm{2}} \right){z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}}\:\Rightarrow \\ $$$$\frac{{z}^{'} }{{z}}\:=\frac{\mathrm{1}}{{x}}\:−{x}\:\Rightarrow\:{ln}\mid{z}\mid\:={ln}\mid{x}\mid\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c}\:\Rightarrow{z}\:={k}\mid{x}\mid\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\:\left({we}\:{search}\:{solutin}\right. \\ $$$${on}\left[\mathrm{0},+\infty\left[\right)\:\Rightarrow{z}\:={k}\:{x}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right. \\ $$$${let}\:{use}\:{mvc}\:{method}\:{we}\:{have}\:{z}^{'} \:={k}^{'} {xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$+{k}\left\{\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:+{x}\left(−{x}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right\}\:={k}^{'} \:{x}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{k}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\left({e}\right)\:\Rightarrow{k}^{'} {x}^{\mathrm{2}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\:{k}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\left({x}^{\mathrm{2}} −\mathrm{1}\right){kxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:={xe}^{−{x}^{\mathrm{2}} } \:\Rightarrow \\ $$$${k}^{'} \:=\frac{{xe}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }\:=\frac{\mathrm{1}}{{x}}\:{e}^{−{x}^{\mathrm{2}} \:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:=\frac{\mathrm{1}}{{x}}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\Rightarrow \\ $$$${k}\left({x}\right)\:=\int_{\mathrm{1}} ^{{x}} \:\frac{{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} }{{t}}\:{dt}\:+{c}\:\:\:\:\left({c}={k}\left(\mathrm{1}\right)\:\:{we}\:{have}\:{z}\left(\mathrm{1}\right)={k}\left(\mathrm{1}\right){e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow\right. \\ $$$${k}\left(\mathrm{1}\right)\:={z}\left(\mathrm{1}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{we}\:{have}\:{z}\left({x}\right)={xk}\left({x}\right){e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$={xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \left\{\:\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} }{{t}}{dt}\:\:+{z}\left(\mathrm{1}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}} \right\}\:\:{we}\:{have}\:{y}^{'} ={z}\:\Rightarrow \\ $$$${y}\left({x}\right)=\int_{.} ^{{x}} \:{z}\left({u}\right){du}\:+{C}\: \\ $$$$=\int_{.} ^{{x}} \:\left\{\:{u}\:{e}^{−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}} \left\{\:\:\:\int_{\mathrm{1}} ^{{u}} \:\frac{{e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} }{{t}}{dt}\:+{z}\left(\mathrm{1}\right)\sqrt{{e}}\right\}\right\}{du}\:\:+{C} \\ $$
Answered by mind is power last updated on 13/Nov/19
$${y}'={z} \\ $$$${homgen} \\ $$$${xz}'+\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}=\mathrm{0} \\ $$$$\Rightarrow\frac{{z}'}{{z}}=\frac{\mathrm{1}}{{x}}−{x}\Rightarrow{ln}\mid{z}\mid=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{ln}\left({x}\right)\Rightarrow{z}={kxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${xz}'+\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}={xe}^{−{x}^{\mathrm{2}} } \\ $$$${x}\left({k}'{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +\left({e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} −{x}^{\mathrm{2}} {e}^{\frac{−{x}^{\mathrm{2}} }{\mathrm{2}}} \right){k}\right)+\left({x}^{\mathrm{2}} −\mathrm{1}\right){kxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} ={xe}^{−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{k}'{x}^{\mathrm{2}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} ={xe}^{−{x}^{\mathrm{2}} } \\ $$$${k}'=\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{{x}} \\ $$$${k}=\int\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}}{dx} \\ $$$${x}=\sqrt{{t}}\Rightarrow{dx}=\frac{{dt}}{\mathrm{2}\sqrt{{t}}} \\ $$$$\int\frac{{e}^{−{t}} }{\mathrm{2}{t}}{dt}=\int\frac{{e}^{{t}} }{\mathrm{2}{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}{E}_{{i}} \left(−{t}\right) \\ $$$${E}_{{i}} ={exponential}\:{integral}\:{function} \\ $$$$\Rightarrow{k}=\frac{{Ei}\left(−{x}^{\mathrm{2}} \right)}{\mathrm{2}}+{c} \\ $$$$\Rightarrow{z}\left({x}\right)={xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \left\{\int\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}}{dx}+{c}\right\}=\frac{\mathrm{1}}{\mathrm{2}}{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {Ei}\left(−{x}^{\mathrm{2}} \right)+{c}\frac{{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{2}} \\ $$$${y}=\int{z}\left({x}\right){dx} \\ $$$${by}\:{part}\:\:=−\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{2}}{Ei}\left(−{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} .\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}}{dx}−\frac{{c}}{\mathrm{4}}{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=−\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{2}}{E}_{{i}} \left(−{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{e}^{−\left(\frac{\mathrm{3}}{\mathrm{2}}\right){x}^{\mathrm{2}} } }{{x}}{dx} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} ={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{e}^{−{t}} \:}{{t}}.{dt}=\frac{\mathrm{1}}{\mathrm{2}}{E}_{{i}} \left(−{t}\right)=\frac{\mathrm{1}}{\mathrm{2}}{E}_{{i}} \left(−\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \right)+{d} \\ $$$${y}\left({x}\right)=−\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{2}}{E}_{{i}} \left(−{x}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{4}}{E}_{{i}} \left(−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right)−\frac{{cxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{4}}+{d} \\ $$$${c},{d}\:{constante} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by arkanmath7@gmail.com last updated on 13/Nov/19
$${let}\:\:{y}^{'} \:=\:{p}\:\:\:,\:\:\:{y}''\:=\:{p}^{'} \\ $$$${xp}\:^{'} \:+\:\left({x}^{\mathrm{2}} \:−\:\mathrm{1}\right){p}\:=\:{xe}^{−{x}^{\mathrm{2}} } \:\:\:\:\:\:\:/{x} \\ $$$${p}^{'} \:\:+\:\left({x}\:−\:\frac{\mathrm{1}}{{x}}\right){p}\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$$${P}\:\left({x}\right)\:=\:{x}\:−\:\frac{\mathrm{1}}{{x}}\:\:\:,\:\:\:{Q}\:\left({x}\right)\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$$${p}\:=\:{e}^{−\int{P}\:\left({x}\right){dx}} \:\left[\int{Q}\:\left({x}\right)\:{e}^{\int{P}\:\left({x}\right){dx}} \:\:{dx}\right] \\ $$$${p}\:=\:{e}^{−\int\left({x}−\frac{\mathrm{1}}{{x}}\right){dx}} \:\left[\int{e}^{−{x}^{\mathrm{2}} } {e}^{\int\left({x}\:−\:\frac{\mathrm{1}}{{x}}\right){dx}} \:\:{dx}\right] \\ $$$${p}\:=\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:{lnx}} \:\left[\int{e}^{−{x}^{\mathrm{2}} } {e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−\:{lnx}} \:\:{dx}\right] \\ $$$${p}\:=\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:{e}^{\:{lnx}} \:\left[\int{e}^{−{x}^{\mathrm{2}} } {e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:} \:{e}^{{lnx}\:} {dx}\right] \\ $$$$ \\ $$$${p}\:=\:{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\left[\int−{x}\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} } {dx}\right] \\ $$$${p}\:=\:{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:\left[\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} } \:+\:{c}\right] \\ $$$${p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{xe}^{−{x}^{\mathrm{2}} } \:\:+\:{cxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{xe}^{−{x}^{\mathrm{2}} } \:\:+\:{cxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$ \\ $$$${dy}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}{xe}^{−{x}^{\mathrm{2}} } \:\:+\:{cxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right){dx} \\ $$$$ \\ $$$$\int{dy}\:=\:\int\left(\frac{\mathrm{1}}{\mathrm{2}}{xe}^{−{x}^{\mathrm{2}} } \:\:+\:{cxe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right){dx} \\ $$$$ \\ $$$${y}\:=\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{x}^{\mathrm{2}} \:} −{ce}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:+\:{c}_{\mathrm{1}} \\ $$
Commented by mind is power last updated on 13/Nov/19
$${lin}\:\mathrm{8}\:{e}^{−{ln}\left({x}\right)} .{not}\:{e}^{{ln}\left({x}\right)} \\ $$
Commented by arkanmath7@gmail.com last updated on 13/Nov/19
$${yes}\:{true} \\ $$
Commented by arkanmath7@gmail.com last updated on 13/Nov/19
$${mistake}! \\ $$