solve (x^2 −1)y^(′′) +(√x)y^′ =xe^(−x)


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Question Number 88423 by abdomathmax last updated on 10/Apr/20

$s o l v e (x^{2} - 1) y^{^{″}} + \sqrt{x} y^{^{'}} = {x e}^{- x}$
Commented by mathmax by abdo last updated on 12/Apr/20
$let y^′ =z (e)⇒(x^2 −1)z^′ +(√x)z =xe^(−x) (he)→(x^2 −1)z^′ +(√x)z =0 ⇒(x^2 −1)z^′ =−(√x)z ⇒ (z^′ /z) =−((√x)/(x^2 −1)) ⇒ln∣z∣ =−∫ ((√x)/(x^2 −1))dx =−(1/2)∫ (√x)((1/(x−1))−(1/(x+1)))dx =(1/2)∫ (((√x)dx)/(x+1))−(1/2)∫ ((√x)/(x−1))dx ∫((√x)/(x+1))dx =_((√x)=t) ∫(t/(t^2 +1))(2t)dt =2 ∫((t^2 +1−1)/(t^2 +1))dt =2t−2arctant +c_1 =2(√x)−2arctan((√x)) +c_1 ∫((√x)/(x−1))dx =_((√x)=t) ∫ ((t(2t)dt)/(t^2 −1)) =2 ∫((t^2 −1+1)/(t^2 −1))dt =2t+∫ ((2dt)/(t^2 −1)) =2t +∫ ((1/(t−1))−(1/(t+1)))dt =2t+ln∣((t+1)/(t−1))∣ +c_2 =2(√x) +ln∣(((√x)+1)/((√x)−1))∣ +c_2 ⇒ ln∣z∣ =(√x)−artan((√x))−(√x)−(1/2)ln∣(((√x)+1)/((√x)−1))∣ +C ⇒ z =K (e^(−arctan((√x))) /(√(∣(((√x)+1)/((√x)−1))∣))) let find solution on{ x/(((√x)+1)/((√x)−1))>0} mvc method ⇒z^′ =K^′ (e^(−arctan((√x))) /(√(((√x)+1)/((√x)−1)))) +K ×((−(1/(2(√x)(1+x)))e^(−arctan((√x))) −e^(−arctan((√x))) ((((((√x)+1)/((√x)−1)))^′ )/(2(√(((√x)+1)/((√x)−1))))))/(((√x)+1)/((√x)−1))) =K^′ (√(((√x)−1)/((√x)+1)))e^(−arctan((√x))) +e^(−arctan((√x))) ×((((√(((√x)+1)/((√x)−1)))/((√x)(1+x)))−((((√x)+1)/((√x)−1)))^′ )/(4(√x)(1+x)(√(((√x)+1)/((√x)−1)))×(((√x)+1)/((√x)−1)))) ...be continued...$
$l e t y^{^{'}} = z (e) \Rightarrow (x^{2} - 1) z^{^{'}} + \sqrt{x} z = {x e}^{- x}$ $(h e) \to (x^{2} - 1) z^{^{'}} + \sqrt{x} z = 0 \Rightarrow (x^{2} - 1) z^{^{'}} = - \sqrt{x} z \Rightarrow$ $\frac{z^{^{'}}}{z} = - \frac{\sqrt{x}}{x^{2} - 1} \Rightarrow l n ∣ z ∣ = - \int \frac{\sqrt{x}}{x^{2} - 1} d x$ $= - \frac{1}{2} \int \sqrt{x} (\frac{1}{x - 1} - \frac{1}{x + 1}) d x = \frac{1}{2} \int \frac{\sqrt{x} d x}{x + 1} - \frac{1}{2} \int \frac{\sqrt{x}}{x - 1} d x$ $\int \frac{\sqrt{x}}{x + 1} d x =_{\sqrt{x} = t} \int \frac{t}{t^{2} + 1} (2 t) d t = 2 \int \frac{t^{2} + 1 - 1}{t^{2} + 1} d t$ $= 2 t - 2 a r c t a n t + c_{1} = 2 \sqrt{x} - 2 a r c t a n (\sqrt{x}) + c_{1}$ $\int \frac{\sqrt{x}}{x - 1} d x =_{\sqrt{x} = t} \int \frac{t (2 t) d t}{t^{2} - 1} = 2 \int \frac{t^{2} - 1 + 1}{t^{2} - 1} d t$ $= 2 t + \int \frac{2 d t}{t^{2} - 1} = 2 t + \int (\frac{1}{t - 1} - \frac{1}{t + 1}) d t$ $= 2 t + l n ∣ \frac{t + 1}{t - 1} ∣ + c_{2} = 2 \sqrt{x} + l n ∣ \frac{\sqrt{x} + 1}{\sqrt{x} - 1} ∣ + c_{2} \Rightarrow$ $l n ∣ z ∣ = \sqrt{x} - a r t a n (\sqrt{x}) - \sqrt{x} - \frac{1}{2} l n ∣ \frac{\sqrt{x} + 1}{\sqrt{x} - 1} ∣ + C \Rightarrow$ $z = K \frac{e^{- a r c t a n (\sqrt{x})}}{\sqrt{∣ \frac{\sqrt{x} + 1}{\sqrt{x} - 1} ∣}} l e t f i n d s o l u t i o n o n {x / \frac{\sqrt{x} + 1}{\sqrt{x} - 1} > 0}$ $m v c m e t h o d \Rightarrow z^{^{'}} = K^{^{'}} \frac{e^{- a r c t a n (\sqrt{x})}}{\sqrt{\frac{\sqrt{x} + 1}{\sqrt{x} - 1}}} + K \times \frac{- \frac{1}{2 \sqrt{x} (1 + x)} e^{- a r c t a n (\sqrt{x})} - e^{- a r c t a n (\sqrt{x})} \frac{{(\frac{\sqrt{x} + 1}{\sqrt{x} - 1})}^{^{'}}}{2 \sqrt{\frac{\sqrt{x} + 1}{\sqrt{x} - 1}}}}{\frac{\sqrt{x} + 1}{\sqrt{x} - 1}}$ $= K^{^{'}} \sqrt{\frac{\sqrt{x} - 1}{\sqrt{x} + 1}} e^{- a r c t a n (\sqrt{x})} + e^{- a r c t a n (\sqrt{x})} \times \frac{\frac{\sqrt{\frac{\sqrt{x} + 1}{\sqrt{x} - 1}}}{\sqrt{x} (1 + x)} - {(\frac{\sqrt{x} + 1}{\sqrt{x} - 1})}^{^{'}}}{4 \sqrt{x} (1 + x) \sqrt{\frac{\sqrt{x} + 1}{\sqrt{x} - 1}} \times \frac{\sqrt{x} + 1}{\sqrt{x} - 1}}$ $. . . b e c o n t i n u e d . . .$
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