find the general solution of 1)y′′′−7y′+6y=x 2)y′′′−3y′′+2y′=(e^x /(1+e^(−x) ))


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Question Number 170329 by ali009 last updated on 21/May/22

$f i n d t h e g e n e r a l s o l u t i o n o f$ $1) y^{‴} - 7 y^{'} + 6 y = x$ $2) y^{‴} - 3 y^{″} + 2 y^{'} = \frac{e^{x}}{1 + e^{- x}}$
	Answered by LEKOUMA last updated on 22/May/22

	$1) y^{^{‴}} - 7 y^{'} + 6 y = x$ $E h$ $y^{‴} - 7 y^{'} + 6 y = 0$ $E c$ $r^{3} - 7 r + 6 = 0$ $(r - 1) (r^{2} + r - 6) = 0$ $r - 1 = 0 o r r^{2} + r - 6 = 0$ $r_{1} = 1 o r r_{2} = - 3, r_{3} = 2$ $y = c_{1} e^{x} + c_{2} e^{- 3 x} + c_{3} e^{2 x}$ $S p$ $y_{p} = x \Rightarrow y_{p} = a x$ $y_{p}^{'} = a$ $y_{p}^{″} = 0$ $y_{p}^{‴} = 0$ $0 - 7 a + 6 a x = x$ $6 a = 1 \Leftrightarrow a = \frac{1}{6}$ $y_{p} = \frac{1}{6} x$ $y (x) = y_{h} + y_{p}$ $y (x) = c_{1} e^{x} + c_{2} e^{- 3 x} + c_{3} e^{2 x} + \frac{1}{6} x$ $c_{1}, c_{2} e t c_{3} c s t e$
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