solve y^(′′) −4y^′ +7y =xe^(−x) sinx


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Question Number 125499 by mathmax by abdo last updated on 11/Dec/20

$solve y^{^{″}} - {4 y}^{^{'}} + 7 y = {xe}^{- x} sinx$
	Answered by mathmax by abdo last updated on 12/Dec/20
	$he)→r^2 −4r+7=0→Δ^′ =4−7=−3 ⇒r_1 =2+i(√3)and r_2 =2−i(√3) ⇒ y_h =ae^((2+i(√3))x) +b e^((2−i(√3))x) =e^(2x) {αcos((√3)x)+βsin((√3)x)} =α e^(2x) cos((√3)x)+β e^(2x) sin((√3)x)=αu_1 +βu_2 W(u_1 ,u_2 )= determinant (((e^(2x) cos((√3)x) e^(2x) sin((√3)x))),(((2cos((√3)x)−(√3)sin((√3)x)e^(2x) (2sin((√3)x)+(√3)cos((√3)x)e^(2x) ))) =e^(4x) { 2cos((√3)x)sin((√3)x)+(√3)cos^2 ((√3)x)}−e^(4x) {2cos((√3)x)sin((√3)x)−(√3)sin^2 ((√3)x)} =e^(4x) ((√3)) =(√3)e^(4x) ≠0 W_1 = determinant (((o e^(2x) sin((√3)x))),((xe^(−x) sinx (2sin((√3)x)+(√3)cos((√3)x))))=−xe^x sinx sin((√3)x) W_2 = determinant (((e^(2x) cos((√3)x) 0)),(((2cos((√3)x)−(√3)sin((√3)x)e^(2x) xe^(−x) sinx))) =xe^x sinx cos((√3)x) v_1 =∫ (w_1 /w)dx =−∫ ((xe^x sinx sin((√3)x))/( (√3)e^(4x) ))=−(1/( (√3)))∫ xe^(−3x) sinxsin((√3)x)dx v_2 = ∫ (w_2 /w)dx =∫ ((xe^x sinx cos((√3)x))/( (√3)e^(4x) ))dx=(1/( (√3)))∫x e^(−3x) sinx cos((√3)x)dx ⇒y_p =u_1 v_1 +u_2 v_2 =−(1/( (√3)))e^(2x) cos((√3)x)∫ x e^(−3x) sinx sin((√3)x)dx +(1/( (√3)))e^(2x) sin((√3)x)∫ xe^(−3x) sinx cos((√3)x)dx and general solution is y =y_h +y_p$
	$he) \to r^{2} - 4 r + 7 = 0 \to Δ^{^{'}} = 4 - 7 = - 3 \Rightarrow r_{1} = 2 + i \sqrt{3} and r_{2} = 2 - i \sqrt{3} \Rightarrow$ $y_{h} = {ae}^{(2 + i \sqrt{3}) x} + b e^{(2 - i \sqrt{3}) x} = e^{2 x} {α \cos (\sqrt{3} x) + β \sin (\sqrt{3} x)}$ $= α e^{2 x} \cos (\sqrt{3} x) + β e^{2 x} \sin (\sqrt{3} x) = α u_{1} + β u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{2 x} \cos (\sqrt{3} x) e^{2 x} \sin (\sqrt{3} x) \\ (2 \cos (\sqrt{3} x) - \sqrt{3} \sin (\sqrt{3} x) e^{2 x} (2 \sin (\sqrt{3} x) + \sqrt{3} \cos (\sqrt{3} x) e^{2 x} \end{matrix} \|$ $= e^{4 x} {2 \cos (\sqrt{3} x) \sin (\sqrt{3} x) + \sqrt{3} \cos^{2} (\sqrt{3} x)} - e^{4 x} {2 \cos (\sqrt{3} x) \sin (\sqrt{3} x) - \sqrt{3} \sin^{2} (\sqrt{3} x)}$ $= e^{4 x} (\sqrt{3}) = \sqrt{3} e^{4 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{2 x} \sin (\sqrt{3} x) \\ {xe}^{- x} sinx (2 \sin (\sqrt{3} x) + \sqrt{3} \cos (\sqrt{3} x) \end{matrix} \| = - {xe}^{x} sinx \sin (\sqrt{3} x)$ $W_{2} = \| \begin{matrix} e^{2 x} \cos (\sqrt{3} x) 0 \\ (2 \cos (\sqrt{3} x) - \sqrt{3} \sin (\sqrt{3} x) e^{2 x} {xe}^{- x} sinx \end{matrix} \|$ $= {xe}^{x} sinx \cos (\sqrt{3} x)$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{{xe}^{x} sinx \sin (\sqrt{3} x)}{\sqrt{3} e^{4 x}} = - \frac{1}{\sqrt{3}} \int {xe}^{- 3 x} sinxsin (\sqrt{3} x) dx$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{{xe}^{x} sinx \cos (\sqrt{3} x)}{\sqrt{3} e^{4 x}} dx = \frac{1}{\sqrt{3}} \int x e^{- 3 x} sinx \cos (\sqrt{3} x) dx$ $\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2}$ $= - \frac{1}{\sqrt{3}} e^{2 x} \cos (\sqrt{3} x) \int x e^{- 3 x} sinx \sin (\sqrt{3} x) dx$ $+ \frac{1}{\sqrt{3}} e^{2 x} \sin (\sqrt{3} x) \int {xe}^{- 3 x} sinx \cos (\sqrt{3} x) dx and general solution is$ $y = y_{h} + y_{p}$
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