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-

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

R e s o l u t i o n

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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 \dots

\dots \dots \dots . . L e p u i s s a n t \dots \dots \dots \dots

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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