solve y^(′′) +2y^′ −y =(e^(−x) /x)


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Question Number 103995 by mathmax by abdo last updated on 18/Jul/20

$solve y^{^{″}} + {2 y}^{^{'}} - y = \frac{e^{- x}}{x}$
	Answered by mathmax by abdo last updated on 20/Jul/20
	$h→r^2 +2r−1 =0 →Δ^′ =1+1=2 ⇒r_1 =−1+(√2) and r_2 =−1−(√2) ⇒ y_h =ae^((−1+(√2))x) )+b e^((−1−(√2))x) =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((e^((−1+(√2))x) e^((−1−(√2))x) )),(((−1+(√2))e^((−1+(√2))x) (−1−(√2)) e^((−1−(√2))x) ))) =(−1−(√2))e^(−2x) −(−1+(√2))e^(−2x) =−2(√2)e^(−2x) W_1 = determinant (((o e^((−1−(√2))x) )),(((e^(−x) /x) (−1−(√2))e^((−1−(√2))x) ))) =−(e^(−x) /x) e^((−1−(√2))x) =−(1/x)e^((−2−(√2))x) W_2 = determinant (((e^((−1+(√2))x) 0)),(((−1+(√2))e^x (e^(−x) /x))))=(1/x) e^((−2+(√2))x) v_1 =∫ (w_1 /w)dx =−∫ (e^((−2+(√2))x) /(−2(√2)x e^(−2x) ))dx =(1/(2(√2))) ∫ (e^((√2)x) /x)dx v_2 =∫ (w_2 /w)dx =−(1/(2(√2)))∫ (e^((−2+(√2))x) /(x e^(−2x) ))dx =−(1/(2(√2)))∫ (e^((√2)x) /x)dx ⇒ y_p = u_1 v_1 +u_2 v_2 =(1/(2(√2)))e^((−1+(√2))x) ∫ (e^(x(√2)) /x)dx −(1/(2(√2)))e^((−1−(√2))x) ∫ (e^(x(√2)) /x)dx =(e^(−x) /(√2)){((e^(x(√2)) −e^(−x(√2)) )/2)}∫ (e^(x(√2)) /2)dx =(e^(−x) /(√2))sh(x(√2))∫ (e^(x(√2)) /x)dx the general solution is y =y_h +y_p ⇒y =a e^((−1+(√2))x) +b e^((−1−(√2))x) + (e^(−x) /(√2))sh(x(√2))∫ (e^(x(√2)) /x)dx$
	$h \to r^{2} + 2 r - 1 = 0 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow r_{1} = - 1 + \sqrt{2} and r_{2} = - 1 - \sqrt{2} \Rightarrow$ $y_{h} = {ae}^{(- 1 + \sqrt{2}) x}) + b e^{(- 1 - \sqrt{2}) x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{(- 1 + \sqrt{2}) x} e^{(- 1 - \sqrt{2}) x} \\ (- 1 + \sqrt{2}) e^{(- 1 + \sqrt{2}) x} (- 1 - \sqrt{2}) e^{(- 1 - \sqrt{2}) x} \end{matrix} \|$ $= (- 1 - \sqrt{2}) e^{- 2 x} - (- 1 + \sqrt{2}) e^{- 2 x} = - 2 \sqrt{2} e^{- 2 x}$ $W_{1} = \| \begin{matrix} o e^{(- 1 - \sqrt{2}) x} \\ \frac{e^{- x}}{x} (- 1 - \sqrt{2}) e^{(- 1 - \sqrt{2}) x} \end{matrix} \| = - \frac{e^{- x}}{x} e^{(- 1 - \sqrt{2}) x} = - \frac{1}{x} e^{(- 2 - \sqrt{2}) x}$ $W_{2} = \| \begin{matrix} e^{(- 1 + \sqrt{2}) x} 0 \\ (- 1 + \sqrt{2}) e^{x} \frac{e^{- x}}{x} \end{matrix} \| = \frac{1}{x} e^{(- 2 + \sqrt{2}) x}$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{e^{(- 2 + \sqrt{2}) x}}{- 2 \sqrt{2} x e^{- 2 x}} dx = \frac{1}{2 \sqrt{2}} \int \frac{e^{\sqrt{2} x}}{x} dx$ $v_{2} = \int \frac{w_{2}}{w} dx = - \frac{1}{2 \sqrt{2}} \int \frac{e^{(- 2 + \sqrt{2}) x}}{x e^{- 2 x}} dx = - \frac{1}{2 \sqrt{2}} \int \frac{e^{\sqrt{2} x}}{x} dx \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = \frac{1}{2 \sqrt{2}} e^{(- 1 + \sqrt{2}) x} \int \frac{e^{x \sqrt{2}}}{x} dx - \frac{1}{2 \sqrt{2}} e^{(- 1 - \sqrt{2}) x} \int \frac{e^{x \sqrt{2}}}{x} dx$ $= \frac{e^{- x}}{\sqrt{2}} {\frac{e^{x \sqrt{2}} - e^{- x \sqrt{2}}}{2}} \int \frac{e^{x \sqrt{2}}}{2} dx = \frac{e^{- x}}{\sqrt{2}} sh (x \sqrt{2}) \int \frac{e^{x \sqrt{2}}}{x} dx$ $the general solution is y = y_{h} + y_{p}$ $\Rightarrow y = a e^{(- 1 + \sqrt{2}) x} + b e^{(- 1 - \sqrt{2}) x} + \frac{e^{- x}}{\sqrt{2}} sh (x \sqrt{2}) \int \frac{e^{x \sqrt{2}}}{x} dx$
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