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) .](https://www.tinkutara.com/question/Q31972.png)
$$\left.{solve}\:{inside}\:\right]−\mathrm{1},\mathrm{1}\left[\:{the}\:{d}.{e}.\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{y}^{'} \:+{y}\:={e}^{−\mathrm{2}{x}} \:.\right. \\ $$
Commented by math khazana by abdo last updated on 15/Aug/18
$${he}\:\Rightarrow\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{y}^{'} \:+{y}\:=\mathrm{0}\:\Rightarrow \\ $$$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{y}^{'} \:=−{y}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int\:\frac{{y}^{'} }{{y}}\:{dx}\:=\:{arccosx}\:+\alpha\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\:{arccosx}\:+\alpha\:\Rightarrow{y}\:={k}\:{e}^{{arccosx}} \\ $$$${mvc}\:{method}\:{give}\:{y}^{'} \:={k}^{'} \:{e}^{{arccosx}} \:−\frac{{k}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{e}^{{arccosx}} \\ $$$$\left({e}\right)\:\Rightarrow\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\left\{{k}^{'} \:{e}^{{arccosx}} \:−\frac{{k}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right\}{e}^{{arccosx}} \\ $$$$+{k}\:{e}^{{arccosx}} \:={e}^{−\mathrm{2}{x}} \:\:\Rightarrow \\ $$$${k}^{'} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{e}^{{arccosx}} \:\:={e}^{−\mathrm{2}{x}} \:\Rightarrow \\ $$$${k}^{'} \:=\:\frac{{e}^{−\mathrm{2}{x}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{e}^{{arccosx}} }\:=\:\frac{{e}^{−\mathrm{2}{x}−{arccosx}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${k}\left({x}\right)\:=\:\int\:\:\:\:\frac{{e}^{−\mathrm{2}{x}\:−{arccosx}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)={e}^{{arcosx}} \:\:\left\{\:\:\int_{.} ^{{x}} \:\:\:\frac{{e}^{−\mathrm{2}{t}\:−{arccost}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+{c}\right\} \\ $$