quel est le changement de variable qui permet de passer de l′equation differentielle : x^2 y′′−3xy′+4y=0 a une equation lineaire d′ordre 2 coefficient comstant en z


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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
	$Resolution −−−−−−−−−−−−−−−−−−−− x^2 y′′ −3xy′+4y=0.. ; posons y=x^m . ⇒ y′=mx^(m−1) ; y′′=m(m−1)x^(m−2) Alors en remplacant dans l′equation, on a : x^2 m(m−1)x^(m−2) −3xmx^(m−1) +4x^m =0 ⇒ m(m−1)x^m −3mx^m +4x^m =0 ⇒ x^m {m(m−1)−3m+4}=0.. ⇒ m(m−1)−3m+4=0 ⇒ m^2 −4m+4=0 → (m−2)^2 =0 ⇒ m=2.. y_1 =x^m =x^2 ; et y_2 =x^m lnx=x^2 lnx.. ∴∵ y=C_1 x^2 +C_2 x^2 lnx.. Divers : Alors, cette equation porte le nom d′equation de Cauchy−Euler... ...........Le puissant............$
	$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Resolution} \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$$${x}^{\mathrm{2}} {y}''\:−\mathrm{3}{xy}'+\mathrm{4}{y}=\mathrm{0}..\:\:;\:\:{posons}\:{y}={x}^{{m}} . \\ $$$$\Rightarrow\:{y}'={mx}^{{m}−\mathrm{1}} \:;\:\:{y}''={m}\left({m}−\mathrm{1}\right){x}^{{m}−\mathrm{2}} \\ $$$${Alors}\:{en}\:{remplacant}\:{dans}\:{l}'{equation}, \\ $$$${on}\:{a}\::\: \\ $$$${x}^{\mathrm{2}} {m}\left({m}−\mathrm{1}\right){x}^{{m}−\mathrm{2}} −\mathrm{3}{xmx}^{{m}−\mathrm{1}} +\mathrm{4}{x}^{{m}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}\left({m}−\mathrm{1}\right){x}^{{m}} −\mathrm{3}{mx}^{{m}} +\mathrm{4}{x}^{{m}} =\mathrm{0} \\ $$$$\Rightarrow\:{x}^{{m}} \left\{{m}\left({m}−\mathrm{1}\right)−\mathrm{3}{m}+\mathrm{4}\right\}=\mathrm{0}.. \\ $$$$\Rightarrow\:{m}\left({m}−\mathrm{1}\right)−\mathrm{3}{m}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:{m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{4}=\mathrm{0}\:\rightarrow\:\left({m}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}=\mathrm{2}.. \\ $$$${y}_{\mathrm{1}} ={x}^{{m}} ={x}^{\mathrm{2}} \:\:;\:\:{et}\:\:{y}_{\mathrm{2}} ={x}^{{m}} {lnx}={x}^{\mathrm{2}} {lnx}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:\:{y}={C}_{\mathrm{1}} {x}^{\mathrm{2}} +{C}_{\mathrm{2}} {x}^{\mathrm{2}} {lnx}.. \\ $$$$ \\ $$$${Divers}\::\:{Alors},\:{cette}\:{equation}\:{porte}\:{le}\:{nom} \\ $$$${d}'{equation}\:{de}\:{Cauchy}−{Euler}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........\mathscr{L}{e}\:{puissant}............ \\ $$
	Commented by SANOGO last updated on 06/Oct/21

	$${toujour}\:{le}\:{puissant} \\ $$
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