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

$$solveinside]-1,1[thed.e.\sqrt{1-x^2}{y}^{{}^{\prime }}+y=e^{-2x}.$$

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

$$he\Rightarrow \sqrt{1-x^2}{y}^{{}^{\prime }}+y=0\Rightarrow$$$$\sqrt{1-x^2}{y}^{{}^{\prime }}=-y\Rightarrow \frac{y^{{}^{\prime }}}{y}=-\frac{1}{\sqrt{1-x^2}}\Rightarrow$$$$\int \frac{y^{{}^{\prime }}}{y}dx=arccosx+\alpha \Rightarrow$$$$ln\mid y\mid =arccosx+\alpha \Rightarrow y=k{e}^{arccosx}$$$$mvcmethodgive{y}^{{}^{\prime }}=k^{{}^{\prime }}{e}^{arccosx}-\frac{k}{\sqrt{1-x^2}}{e}^{arccosx}$$$$\left(e\right)\Rightarrow \sqrt{1-x^2}\{k^{{}^{\prime }}{e}^{arccosx}-\frac{k}{\sqrt{1-x^2}}\}{e}^{arccosx}$$$$+k{e}^{arccosx}=e^{-2x}\Rightarrow$$$$k^{{}^{\prime }}\sqrt{1-x^2}{e}^{arccosx}=e^{-2x}\Rightarrow$$$$k^{{}^{\prime }}=\frac{e^{-2x}}{\sqrt{1-x^2}{e}^{arccosx}}=\frac{e^{-2x-arccosx}}{\sqrt{1-x^2}}\Rightarrow$$$$k\left(x\right)=\int \frac{e^{-2x-arccosx}}{\sqrt{1-x^2}}dx+c\Rightarrow$$$$y\left(x\right)=e^{arcosx}\{{\int }_{.}^x\frac{e^{-2t-arccost}}{\sqrt{1-t^2}}+c\}$$

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