solve y^(′′) −y^′ +y =(e^(−x) /(x+1))


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Question Number 131048 by mathmax by abdo last updated on 31/Jan/21

$solve y^{^{″}} - y^{^{'}} + y = \frac{e^{- x}}{x + 1}$
	Answered by Ar Brandon last updated on 01/Feb/21
	$Homogenous eqn: m^2 −m+1=0 ⇒m=((1±i(√3))/2) y_(gh) =e^(x/2) (Acos(√3)x+Bsin(√3)x) By varying parameters, let y_(PI) =e^(x/2) (A(x)cos(√3)x+B(x)sin(√3)x)=au+bv We solve for a^′ and b′ in the simultaneous eqn below { ((a′u+b′v=0)),((a′u′+b′v′=(e^(−x) /(x+1)))) :} Let W(u,v)= determinant ((u,v),((u′),(v′))) = determinant (((e^(x/2) cos(√3)x),(e^(x/2) sin(√3)x)),(((1/2)e^(x/2) cos(√3)x−(√3)e^(x/2) sin(√3)x),((√3)e^(x/2) cos(√3)x+(1/2)e^(x/2) sin(√3)x))) =e^x ((√3)cos^2 (√3)x+(1/2)sin(√3)xcos(√3)x) −e^x ((1/2)sin(√3)xcos(√3)x−(√3)sin^2 (√3)x)=(√3)e^x w_u = determinant ((0,(e^(x/2) sin(√3)x)),((e^(−x) /(x+1)),((√3)e^(x/2) cos(√3)x+(1/2)e^(x/2) sin(√3)x)))=−((e^(−(x/2)) sin(√3)x)/(x+1)) w_v = determinant (((e^(x/2) cos(√3)x),0),(((1/2)e^(x/2) cos(√3)x−(√3)e^(x/2) sin(√3)x),(e^(−x) /(x+1))))=((e^(−(x/2)) cos(√3)x)/(x+1)) u=∫(w_u /W)dx , v=∫(w_v /W)dx$
	$Homogenous eqn : m^{2} - m + 1 = 0 \Rightarrow m = \frac{1 \pm i \sqrt{3}}{2}$ $y_{gh} = e^{\frac{x}{2}} (Acos \sqrt{3} x + Bsin \sqrt{3} x)$ $By varying parameters, let$ $y_{PI} = e^{\frac{x}{2}} (A (x) \cos \sqrt{3} x + B (x) \sin \sqrt{3} x) = au + bv$ $We solve for a^{^{'}} and b^{'} in the simultaneous eqn below$ ${\begin{cases} a^{'} u + b^{'} v = 0 \\ a^{'} u^{'} + b^{'} v^{'} = \frac{e^{- x}}{x + 1} \end{cases}$ $Let W (u, v) = \| \begin{matrix} u & v \\ u^{'} & v^{'} \end{matrix} \|$ $= \| \begin{matrix} e^{\frac{x}{2}} \cos \sqrt{3} x & e^{\frac{x}{2}} \sin \sqrt{3} x \\ \frac{1}{2} e^{\frac{x}{2}} \cos \sqrt{3} x - \sqrt{3} e^{\frac{x}{2}} \sin \sqrt{3} x & \sqrt{3} e^{\frac{x}{2}} \cos \sqrt{3} x + \frac{1}{2} e^{\frac{x}{2}} \sin \sqrt{3} x \end{matrix} \|$ $= e^{x} (\sqrt{3} \cos^{2} \sqrt{3} x + \frac{1}{2} \sin \sqrt{3} xcos \sqrt{3} x)$ $- e^{x} (\frac{1}{2} \sin \sqrt{3} xcos \sqrt{3} x - \sqrt{3} \sin^{2} \sqrt{3} x) = \sqrt{3} e^{x}$ $w_{u} = \| \begin{matrix} 0 & e^{\frac{x}{2}} \sin \sqrt{3} x \\ \frac{e^{- x}}{x + 1} & \sqrt{3} e^{\frac{x}{2}} \cos \sqrt{3} x + \frac{1}{2} e^{\frac{x}{2}} \sin \sqrt{3} x \end{matrix} \| = - \frac{e^{- \frac{x}{2}} \sin \sqrt{3} x}{x + 1}$ $w_{v} = \| \begin{matrix} e^{\frac{x}{2}} \cos \sqrt{3} x & 0 \\ \frac{1}{2} e^{\frac{x}{2}} \cos \sqrt{3} x - \sqrt{3} e^{\frac{x}{2}} \sin \sqrt{3} x & \frac{e^{- x}}{x + 1} \end{matrix} \| = \frac{e^{- \frac{x}{2}} \cos \sqrt{3} x}{x + 1}$ $u = \int \frac{w_{u}}{W} dx, v = \int \frac{w_{v}}{W} dx$
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