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

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

$${solve}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}^{''} \:+\sqrt{{x}}{y}^{'} \:={xe}^{−{x}} \\ $$

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...

$${let}\:\:{y}^{'} \:={z}\:\:\:\left({e}\right)\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}^{'} \:+\sqrt{{x}}{z}\:={xe}^{−{x}} \\ $$$$\left({he}\right)\rightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}^{'} \:+\sqrt{{x}}{z}\:=\mathrm{0}\:\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}^{'} \:=−\sqrt{{x}}{z}\:\Rightarrow \\ $$$$\frac{{z}^{'} }{{z}}\:=−\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{ln}\mid{z}\mid\:=−\int\:\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{{x}}\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\sqrt{{x}}{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\sqrt{{x}}}{{x}−\mathrm{1}}{dx} \\ $$$$\int\frac{\sqrt{{x}}}{{x}+\mathrm{1}}{dx}\:=_{\sqrt{{x}}={t}} \:\:\:\int\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\frac{{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\mathrm{2}{t}−\mathrm{2}{arctant}\:+{c}_{\mathrm{1}} =\mathrm{2}\sqrt{{x}}−\mathrm{2}{arctan}\left(\sqrt{{x}}\right)\:+{c}_{\mathrm{1}} \\ $$$$\int\frac{\sqrt{{x}}}{{x}−\mathrm{1}}{dx}\:=_{\sqrt{{x}}={t}} \:\:\int\:\frac{{t}\left(\mathrm{2}{t}\right){dt}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\mathrm{2}\:\int\frac{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$=\mathrm{2}{t}+\int\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\mathrm{2}{t}\:+\int\:\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\mathrm{2}{t}+{ln}\mid\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\mid\:+{c}_{\mathrm{2}} =\mathrm{2}\sqrt{{x}}\:+{ln}\mid\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}\mid\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\sqrt{{x}}−{artan}\left(\sqrt{{x}}\right)−\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}\mid\:+{C}\:\Rightarrow \\ $$$${z}\:={K}\:\:\frac{{e}^{−{arctan}\left(\sqrt{{x}}\right)} }{\sqrt{\mid\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}\mid}}\:{let}\:{find}\:{solution}\:{on}\left\{\:{x}/\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}>\mathrm{0}\right\} \\ $$$${mvc}\:{method}\:\Rightarrow{z}^{'} \:={K}\:^{'} \:\frac{{e}^{−{arctan}\left(\sqrt{{x}}\right)} }{\sqrt{\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}}}\:+{K}\:×\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}\left(\mathrm{1}+{x}\right)}{e}^{−{arctan}\left(\sqrt{{x}}\right)} −{e}^{−{arctan}\left(\sqrt{{x}}\right)} \frac{\left(\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}\right)^{'} }{\mathrm{2}\sqrt{\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}}}}{\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}} \\ $$$$={K}\:^{'} \:\sqrt{\frac{\sqrt{{x}}−\mathrm{1}}{\sqrt{{x}}+\mathrm{1}}}{e}^{−{arctan}\left(\sqrt{{x}}\right)} \:+{e}^{−{arctan}\left(\sqrt{{x}}\right)} ×\frac{\frac{\sqrt{\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}}}{\sqrt{{x}}\left(\mathrm{1}+{x}\right)}−\left(\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}\right)^{'} }{\mathrm{4}\sqrt{{x}}\left(\mathrm{1}+{x}\right)\sqrt{\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}}×\frac{\sqrt{{x}}+\mathrm{1}}{\sqrt{{x}}−\mathrm{1}}} \\ $$$$...{be}\:{continued}... \\ $$

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