Question Number 159346 by Ar Brandon last updated on 15/Nov/21
$${xy}''+\mathrm{2}\left({x}+\mathrm{1}\right){y}'+\left({x}+\mathrm{2}\right){y}=\mathrm{0} \\ $$
Answered by mindispower last updated on 16/Nov/21
$${x}\left({y}''+\mathrm{2}{y}'+{y}\right)+\left({y}'+{y}\right) \\ $$$$={x}\left(\left({y}'+{y}\right)'+\left({y}'+{y}\right)\right)+\left({y}'+{y}\right)=\mathrm{0} \\ $$$${z}={y}+{y}' \\ $$$$\Rightarrow{x}\left({z}'+{z}\right)+{z}=\mathrm{0} \\ $$$$\Rightarrow\frac{{z}'}{{z}}=\left(−\frac{\mathrm{1}}{{x}}−\mathrm{1}\right) \\ $$$${ln}\mid{z}\mid=−{x}−{ln}\left({x}\right)+{c} \\ $$$$\Rightarrow{z}={k}\frac{{e}^{−{x}} }{{x}} \\ $$$${y}'+{y}={k}\frac{{e}^{−{x}} }{{x}} \\ $$$${y}'+{y}=\mathrm{0}\Rightarrow{y}={ce}^{−{x}} \\ $$$${y}={c}\left({x}\right){e}^{−{x}} \:\: \\ $$$${c}'={k}\frac{\mathrm{1}}{{x}} \\ $$$${c}={kln}\left({x}\right)+{a} \\ $$$${y}\left({x}\right)=\left({k}\:{ln}\left({x}\right)+{a}\right){e}^{−{x}} \\ $$
Commented by Ar Brandon last updated on 16/Nov/21
Merci monsieur ��
Commented by mindispower last updated on 16/Nov/21
$${je}\:{t}\:{en}\:{prie}\: \\ $$$${je}\left[{vais}\:{faire}\:{une}\:{email}\:{secondaire}\:{je}\:{te}\left[{le}\:{passe}\:{apres}\:\right.\right. \\ $$
Commented by Ar Brandon last updated on 17/Nov/21
D'accord monsieur