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Question Number 31972 by abdo imad last updated on 17/Mar/18

solve inside ]−1,1[ the d.e. (√(1−x^2 )) y^′  +y =e^(−2x)  .

$$\left.{solve}\:{inside}\:\right]−\mathrm{1},\mathrm{1}\left[\:{the}\:{d}.{e}.\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{y}^{'} \:+{y}\:={e}^{−\mathrm{2}{x}} \:.\right. \\ $$

Commented by math khazana by abdo last updated on 15/Aug/18

he ⇒(√(1−x^2 )) y^′  +y =0 ⇒  (√(1−x^2 ))y^′  =−y ⇒(y^′ /y) =−(1/(√(1−x^2 ))) ⇒  ∫ (y^′ /y) dx = arccosx +α ⇒  ln∣y∣ = arccosx +α ⇒y =k e^(arccosx)   mvc method give y^′  =k^′  e^(arccosx)  −(k/(√(1−x^2 )))e^(arccosx)   (e) ⇒(√(1−x^2 )){k^′  e^(arccosx)  −(k/(√(1−x^2 )))}e^(arccosx)   +k e^(arccosx)  =e^(−2x)   ⇒  k^′ (√(1−x^2 ))e^(arccosx)   =e^(−2x)  ⇒  k^′  = (e^(−2x) /((√(1−x^2 ))e^(arccosx) )) = (e^(−2x−arccosx) /(√(1−x^2 ))) ⇒  k(x) = ∫    (e^(−2x −arccosx) /(√(1−x^2 ))) dx+c ⇒  y(x)=e^(arcosx)   {  ∫_. ^x    (e^(−2t −arccost) /(√(1−t^2 ))) +c}

$${he}\:\Rightarrow\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{y}^{'} \:+{y}\:=\mathrm{0}\:\Rightarrow \\ $$$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{y}^{'} \:=−{y}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int\:\frac{{y}^{'} }{{y}}\:{dx}\:=\:{arccosx}\:+\alpha\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\:{arccosx}\:+\alpha\:\Rightarrow{y}\:={k}\:{e}^{{arccosx}} \\ $$$${mvc}\:{method}\:{give}\:{y}^{'} \:={k}^{'} \:{e}^{{arccosx}} \:−\frac{{k}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{e}^{{arccosx}} \\ $$$$\left({e}\right)\:\Rightarrow\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\left\{{k}^{'} \:{e}^{{arccosx}} \:−\frac{{k}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right\}{e}^{{arccosx}} \\ $$$$+{k}\:{e}^{{arccosx}} \:={e}^{−\mathrm{2}{x}} \:\:\Rightarrow \\ $$$${k}^{'} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{e}^{{arccosx}} \:\:={e}^{−\mathrm{2}{x}} \:\Rightarrow \\ $$$${k}^{'} \:=\:\frac{{e}^{−\mathrm{2}{x}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{e}^{{arccosx}} }\:=\:\frac{{e}^{−\mathrm{2}{x}−{arccosx}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${k}\left({x}\right)\:=\:\int\:\:\:\:\frac{{e}^{−\mathrm{2}{x}\:−{arccosx}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)={e}^{{arcosx}} \:\:\left\{\:\:\int_{.} ^{{x}} \:\:\:\frac{{e}^{−\mathrm{2}{t}\:−{arccost}} }{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+{c}\right\} \\ $$

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