soit:y′′−3y′−4y=3e^(3x ) avec , f(o)=−(1/2) et f′(0)=4 alors f(1)=?


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 154476 by SANOGO last updated on 18/Sep/21

$s o i t : y^{″} - 3 y^{'} - 4 y = 3 e^{3 x} a v e c,$ $f (o) = - \frac{1}{2} e t f^{'} (0) = 4$ $a l o r s f (1) = ?$
	Answered by ARUNG_Brandon_MBU last updated on 18/Sep/21
	$y′′−3y′−4y=3e^(3x) Solution de l′e^� quation ge^� ne^� rale homoge^� ne; m^2 −3m−4=0⇒(m−4)(m+1)=0 y_(gh) =ae^(4x) +be^(−x) . Par la mvc posons y_p =a(x)e^(4x) +b(x)e^(−x) { ((a′(x)e^(4x) +b′(x)e^(−x) =0)),((4a′(x)e^(4x) −b′(x)e^(−x) =3e^(3x) )) :} W= determinant ((e^(4x) ,e^(−x) ),((4e^(4x) ),(−e^(−x) )))=−e^(3x) −4e^(3x) =−5e^(3x) ≠0 W_1 = determinant ((0,e^(−x) ),((3e^(3x) ),(−e^(−x) )))=−3e^(2x) , W_2 = determinant ((e^(4x) ,0),((4e^(4x) ),(3e^(3x) )))=3e^(7x) a(x)=∫(W_1 /W)dx=(3/5)∫e^(−x) dx=−(3/5)e^(−x) +C b(x)=∫(W_2 /W)dx=−(3/5)∫e^(4x) dx=−(3/(20))e^(4x) +K Y_G =y_(gh) +y_p =ae^(4x) +be^(−x) −(3/5)e^(3x) −(3/(20))e^(3x) f(0)=a+b−(3/4)=−(1/2)⇒a+b=(1/4) f ′(0)=4a−b−(9/5)−(9/(20))=4⇒4a−b=((25)/4) 1−5b=((25)/4), b=−((21)/(20)), a=((13)/(10)) f(x)=((13)/(10))e^(4x) −((21)/(20))e^(−x) −(3/4)e^(3x)$
	$y^{″} - {3 y}^{'} - 4 y = {3 e}^{3 x}$ $Solution de l^{'} \overset{´}{e q u a t i o n} g \overset{´}{e n} \overset{´}{e r a l e} homog \overset{`}{e n e};$ $m^{2} - 3 m - 4 = 0 \Rightarrow (m - 4) (m + 1) = 0$ $y_{gh} = {ae}^{4 x} + {be}^{- x} .$ $Par la mvc posons y_{p} = a (x) e^{4 x} + b (x) e^{- x}$ ${\begin{cases} a^{'} (x) e^{4 x} + b^{'} (x) e^{- x} = 0 \\ {4 a}^{'} (x) e^{4 x} - b^{'} (x) e^{- x} = {3 e}^{3 x} \end{cases}$ $W = \| \begin{matrix} e^{4 x} & e^{- x} \\ {4 e}^{4 x} & - e^{- x} \end{matrix} \| = - e^{3 x} - {4 e}^{3 x} = - {5 e}^{3 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 & e^{- x} \\ {3 e}^{3 x} & - e^{- x} \end{matrix} \| = - {3 e}^{2 x}, W_{2} = \| \begin{matrix} e^{4 x} & 0 \\ {4 e}^{4 x} & {3 e}^{3 x} \end{matrix} \| = {3 e}^{7 x}$ $a (x) = \int \frac{W_{1}}{W} dx = \frac{3}{5} \int e^{- x} dx = - \frac{3}{5} e^{- x} + C$ $b (x) = \int \frac{W_{2}}{W} dx = - \frac{3}{5} \int e^{4 x} dx = - \frac{3}{20} e^{4 x} + K$ $Y_{G} = y_{gh} + y_{p} = {ae}^{4 x} + {be}^{- x} - \frac{3}{5} e^{3 x} - \frac{3}{20} e^{3 x}$ $f (0) = a + b - \frac{3}{4} = - \frac{1}{2} \Rightarrow a + b = \frac{1}{4}$ $f^{'} (0) = 4 a - b - \frac{9}{5} - \frac{9}{20} = 4 \Rightarrow 4 a - b = \frac{25}{4}$ $1 - 5 b = \frac{25}{4}, b = - \frac{21}{20}, a = \frac{13}{10}$ $f (x) = \frac{13}{10} e^{4 x} - \frac{21}{20} e^{- x} - \frac{3}{4} e^{3 x}$
	Commented by SANOGO last updated on 18/Sep/21

	$m e r c i b i e n l e d u r$
	Commented by ARUNG_Brandon_MBU last updated on 18/Sep/21

	$Autrement posons y_{p} = k e^{3 x}$
	Commented by SANOGO last updated on 18/Sep/21

	$o k d a c c o r d m e r c i$
	Commented by Ar Brandon last updated on 18/Sep/21

	$Je vous en prie$
	Answered by qaz last updated on 19/Sep/21

	$y_{p} = \frac{1}{D^{2} - 3 D - 4} {3 e}^{3 x} = 3 \cdot \frac{1}{3^{2} - 3 \cdot 3 - 4} e^{3 x} = - \frac{3}{4} e^{3 x}$ $\Rightarrow y = C_{1} e^{- x} + C_{2} e^{4 x} - \frac{3}{4} e^{3 x}$ $y^{'} (x) = - C_{1} e^{- x} + {4 C}_{2} e^{4 x} - \frac{9}{4} e^{3 x}$ $y (0) = - 1 / 2 y^{'} (0) = 4$ $\Rightarrow C_{1} = - \frac{21}{20} C_{2} = \frac{13}{10}$ $\Rightarrow y = - \frac{21}{20} e^{- x} + \frac{13}{10} e^{4 x} - \frac{3}{4} e^{3 x}$
	Answered by physicstutes last updated on 20/Sep/21

	$auxillary equation : m^{2} - 3 m - 4 = 0$ $\Rightarrow (m + 1) (m - 4) = 0$ $m = - 1 or m = 4$ $y = {A e}^{- x} + {B e}^{4 x}, A, B \in R$ $f (0) = - \frac{1}{2} \Rightarrow - \frac{1}{2} = A + B . . . . (i)$ $f^{'} (x) = - {A e}^{- x} + 4 B^{4 x}$ $\Rightarrow 4 = - A + 4 B . . . . . (i i)$ $\frac{7}{2} = 5 B \Rightarrow B = \frac{7}{10}$ $A = - \frac{1}{2} - \frac{7}{10} = - \frac{6}{5}$ $\Rightarrow y = - \frac{6}{5} e^{- x} + \frac{7}{10} e^{4 x}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum