Question Number 155959 by SANOGO last updated on 06/Oct/21
$${quel}\:{est}\:{le}\:{changement}\:{de}\:{variable}\:{qui}\:{permet}\:{de}\:{passer} \\ $$$${de}\:{l}'{equation}\:{differentielle}\:: \\ $$$${x}^{\mathrm{2}} {y}''−\mathrm{3}{xy}'+\mathrm{4}{y}=\mathrm{0} \\ $$$${a}\:{une}\:{equation}\:{lineaire}\:{d}'{ordre}\:\mathrm{2}\:\:{coefficient}\:{comstant}\:\:{en}\:{z} \\ $$
Answered by puissant last updated on 06/Oct/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Resolution} \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$${x}^{\mathrm{2}} {y}''\:−\mathrm{3}{xy}'+\mathrm{4}{y}=\mathrm{0}..\:\:;\:\:{posons}\:{y}={x}^{{m}} . \\ $$$$\Rightarrow\:{y}'={mx}^{{m}−\mathrm{1}} \:;\:\:{y}''={m}\left({m}−\mathrm{1}\right){x}^{{m}−\mathrm{2}} \\ $$$${Alors}\:{en}\:{remplacant}\:{dans}\:{l}'{equation}, \\ $$$${on}\:{a}\::\: \\ $$$${x}^{\mathrm{2}} {m}\left({m}−\mathrm{1}\right){x}^{{m}−\mathrm{2}} −\mathrm{3}{xmx}^{{m}−\mathrm{1}} +\mathrm{4}{x}^{{m}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}\left({m}−\mathrm{1}\right){x}^{{m}} −\mathrm{3}{mx}^{{m}} +\mathrm{4}{x}^{{m}} =\mathrm{0} \\ $$$$\Rightarrow\:{x}^{{m}} \left\{{m}\left({m}−\mathrm{1}\right)−\mathrm{3}{m}+\mathrm{4}\right\}=\mathrm{0}.. \\ $$$$\Rightarrow\:{m}\left({m}−\mathrm{1}\right)−\mathrm{3}{m}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:{m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{4}=\mathrm{0}\:\rightarrow\:\left({m}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}=\mathrm{2}.. \\ $$$${y}_{\mathrm{1}} ={x}^{{m}} ={x}^{\mathrm{2}} \:\:;\:\:{et}\:\:{y}_{\mathrm{2}} ={x}^{{m}} {lnx}={x}^{\mathrm{2}} {lnx}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:\:{y}={C}_{\mathrm{1}} {x}^{\mathrm{2}} +{C}_{\mathrm{2}} {x}^{\mathrm{2}} {lnx}.. \\ $$$$ \\ $$$${Divers}\::\:{Alors},\:{cette}\:{equation}\:{porte}\:{le}\:{nom} \\ $$$${d}'{equation}\:{de}\:{Cauchy}−{Euler}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………..\mathscr{L}{e}\:{puissant}………… \\ $$
Commented by SANOGO last updated on 06/Oct/21
$${toujour}\:{le}\:{puissant} \\ $$