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

$q u e l e s t l e c h a n g e m e n t d e v a r i a b l e q u i p e r m e t d e p a s s e r$ $d e l^{'} e q u a t i o n d i f f e r e n t i e l l e :$ $x^{2} y^{″} - 3 {x y}^{'} + 4 y = 0$ $a u n e e q u a t i o n l i n e a i r e d^{'} o r d r e 2 c o e f f i c i e n t c o m s t a n t e n z$
	Answered by puissant last updated on 06/Oct/21
	$Resolution −−−−−−−−−−−−−−−−−−−− x^2 y′′ −3xy′+4y=0.. ; posons y=x^m . ⇒ y′=mx^(m−1) ; y′′=m(m−1)x^(m−2) Alors en remplacant dans l′equation, on a : x^2 m(m−1)x^(m−2) −3xmx^(m−1) +4x^m =0 ⇒ m(m−1)x^m −3mx^m +4x^m =0 ⇒ x^m {m(m−1)−3m+4}=0.. ⇒ m(m−1)−3m+4=0 ⇒ m^2 −4m+4=0 → (m−2)^2 =0 ⇒ m=2.. y_1 =x^m =x^2 ; et y_2 =x^m lnx=x^2 lnx.. ∴∵ y=C_1 x^2 +C_2 x^2 lnx.. Divers : Alors, cette equation porte le nom d′equation de Cauchy−Euler... ...........Le puissant............$
	$R e s o l u t i o n$ $- - - - - - - - - - - - - - - - - - - -$ $x^{2} y^{″} - 3 {x y}^{'} + 4 y = 0 . .; p o s o n s y = x^{m} .$ $\Rightarrow y^{'} = {m x}^{m - 1}; y^{″} = m (m - 1) x^{m - 2}$ $A l o r s e n r e m p l a c a n t d a n s l^{'} e q u a t i o n,$ $o n a :$ $x^{2} m (m - 1) x^{m - 2} - 3 {x m x}^{m - 1} + 4 x^{m} = 0$ $\Rightarrow m (m - 1) x^{m} - 3 {m x}^{m} + 4 x^{m} = 0$ $\Rightarrow x^{m} {m (m - 1) - 3 m + 4} = 0 . .$ $\Rightarrow m (m - 1) - 3 m + 4 = 0$ $\Rightarrow m^{2} - 4 m + 4 = 0 \to {(m - 2)}^{2} = 0$ $\Rightarrow m = 2 . .$ $y_{1} = x^{m} = x^{2}; e t y_{2} = x^{m} l n x = x^{2} l n x . .$ $∴∵ y = C_{1} x^{2} + C_{2} x^{2} l n x . .$ $D i v e r s : A l o r s, c e t t e e q u a t i o n p o r t e l e n o m$ $d^{'} e q u a t i o n d e C a u c h y - E u l e r . . .$ $. . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . .$
	Commented by SANOGO last updated on 06/Oct/21

	$t o u j o u r l e p u i s s a n t$
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