y′′−y = e^(−2x) sin (e^(−x) )


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Question Number 131086 by EDWIN88 last updated on 01/Feb/21

$y^{″} - y = e^{- 2 x} \sin (e^{- x})$
	Answered by liberty last updated on 01/Feb/21

	$[D^{2} - 1] y = e^{- 2 x} \sin (e^{- x})$ $(D - 1) (D + 1) y = e^{- 2 x} \sin (e^{- x})$ $(D - 1) [e^{x} (D + 1) y] = e^{- x} \sin (e^{- x})$ $(D - 1) D (e^{x} y) = e^{- x} \sin (e^{- x})$ $(D - 1) e^{x} y = \int e^{- x} \sin (e^{- x}) dx$ $(D - 1) e^{x} y = \cos (e^{- x}) + C_{1}$ $(D - 1) y = e^{- x} \cos (e^{- x}) + C_{1} e^{- x}$ $(D - 1) e^{- x} y = e^{- 2 x} + C_{1} e^{- 2 x}$ $D (e^{- x} y) = e^{- 2 x} + C_{1} e^{- 2 x}$ $e^{- x} y = \int (e^{- 2 x} + C_{1} e^{- 2 x}) dx$ $y = e^{x} [- e^{- x} \sin (e^{- x}) - \cos (e^{- x}) - \frac{1}{2} C_{1} e^{- 2 x} + C_{2}]$
	Commented by EDWIN88 last updated on 01/Feb/21

	$w a w . . .$
	Commented by EDWIN88 last updated on 01/Feb/21
	զարմանալի
	Answered by Ar Brandon last updated on 01/Feb/21
	$y′′−y=e^(−2x) sin(e^(−x) ) For homogenous equation m^2 −1=0 ⇒m=1, m=−1 ⇒y_(gh) =Ae^x +Be^(−x) By varying parameters, let y_(PI) =A(x)e^x +B(x)e^(−x) ⇒y_(PI) =au+bv. Solve for a′ and b′ below { ((a′u+b′v=0)),((a′u′+b′v′=e^(−2x) sin(e^(−x) ))) :} W(u, v)= determinant ((u,v),((u′),(v′)))= determinant ((e^x ,e^(−x) ),(e^x ,(−e^(−x) )))=−2≠0 W_u = determinant ((0,e^(−x) ),((e^(−2x) sin(e^(−x) )),(−e^(−x) )))=−e^(−3x) sin(e^(−x) ) W_v = determinant ((e^x ,0),(e^x ,(e^(−2x) sin(e^(−x) ))))=e^(−x) sin(e^(−x) ) a=∫(W_u /W)dx=(1/2)∫e^(−3x) sin(e^(−x) )dx=−(1/2)∫p^2 sin(p)dp, p=e^(−x) =−(1/2){−p^2 cos(p)+2∫pcos(p)dp} =(1/2)p^2 cos(p)−{psin(p)−∫sin(p)dp} =(1/2)e^(−2x) cos(e^(−x) )−e^(−x) sin(e^(−x) )−cos(e^(−x) )+C_1 b=∫(W_v /W)dx=−(1/2)∫e^(−x) sin(e^(−x) )dx=(1/2)∫sin(q)dq =−(1/2)cos(e^(−x) )+C_2 y_(PI) =(1/2)e^(−x) cos(e^(−x) )−sin(e^(−x) )−e^x cos(e^(−x) )+C_1 e^x −(1/2)e^(−x) cos(e^(−x) )+C_2 e^(−x) Y=y_(gh) +y_(PI) =αe^x +βe^(−x) −sin(e^(−x) )−e^x cos(e^(−x) )$
	$y^{″} - y = e^{- 2 x} \sin (e^{- x})$ $For homogenous equation$ $m^{2} - 1 = 0 \Rightarrow m = 1, m = - 1$ $\Rightarrow y_{gh} = {Ae}^{x} + {Be}^{- x}$ $By varying parameters, let y_{PI} = A (x) e^{x} + B (x) e^{- x}$ $\Rightarrow y_{PI} = au + bv . Solve for a^{'} and b^{'} below$ ${\begin{cases} a^{'} u + b^{'} v = 0 \\ a^{'} u^{'} + b^{'} v^{'} = e^{- 2 x} \sin (e^{- x}) \end{cases}$ $W (u, v) = \| \begin{matrix} u & v \\ u^{'} & v^{'} \end{matrix} \| = \| \begin{matrix} e^{x} & e^{- x} \\ e^{x} & - e^{- x} \end{matrix} \| = - 2 \neq 0$ $W_{u} = \| \begin{matrix} 0 & e^{- x} \\ e^{- 2 x} \sin (e^{- x}) & - e^{- x} \end{matrix} \| = - e^{- 3 x} \sin (e^{- x})$ $W_{v} = \| \begin{matrix} e^{x} & 0 \\ e^{x} & e^{- 2 x} \sin (e^{- x}) \end{matrix} \| = e^{- x} \sin (e^{- x})$ $a = \int \frac{W_{u}}{W} dx = \frac{1}{2} \int e^{- 3 x} \sin (e^{- x}) dx = - \frac{1}{2} \int p^{2} \sin (p) dp, p = e^{- x}$ $= - \frac{1}{2} {- p^{2} \cos (p) + 2 \int pcos (p) dp}$ $= \frac{1}{2} p^{2} \cos (p) - {psin (p) - \int \sin (p) dp}$ $= \frac{1}{2} e^{- 2 x} \cos (e^{- x}) - e^{- x} \sin (e^{- x}) - \cos (e^{- x}) + C_{1}$ $b = \int \frac{W_{v}}{W} dx = - \frac{1}{2} \int e^{- x} \sin (e^{- x}) dx = \frac{1}{2} \int \sin (q) dq$ $= - \frac{1}{2} \cos (e^{- x}) + C_{2}$ $y_{PI} = \frac{1}{2} e^{- x} \cos (e^{- x}) - \sin (e^{- x}) - e^{x} \cos (e^{- x}) + C_{1} e^{x}$ $- \frac{1}{2} e^{- x} \cos (e^{- x}) + C_{2} e^{- x}$ $Y = y_{gh} + y_{PI}$ $= α e^{x} + β e^{- x} - \sin (e^{- x}) - e^{x} \cos (e^{- x})$
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