solve-inside-1-1-the-d-e-1-x-2-y-y-e-2x-

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

s o l v e i n s i d e] - 1, 1 [t h e d . e . \sqrt{1 - x^{2}} y^{^{'}} + y = e^{- 2 x} .

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

h e \Rightarrow \sqrt{1 - x^{2}} y^{^{'}} + y = 0 \Rightarrow

\sqrt{1 - x^{2}} y^{^{'}} = - y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{1}{\sqrt{1 - x^{2}}} \Rightarrow

\int \frac{y^{^{'}}}{y} d x = a r c c o s x + α \Rightarrow

l n ∣ y ∣ = a r c c o s x + α \Rightarrow y = k e^{a r c c o s x}

m v c m e t h o d g i v e y^{^{'}} = k^{^{'}} e^{a r c c o s x} - \frac{k}{\sqrt{1 - x^{2}}} e^{a r c c o s x}

(e) \Rightarrow \sqrt{1 - x^{2}} {k^{^{'}} e^{a r c c o s x} - \frac{k}{\sqrt{1 - x^{2}}}} e^{a r c c o s x}

+ k e^{a r c c o s x} = e^{- 2 x} \Rightarrow

k^{^{'}} \sqrt{1 - x^{2}} e^{a r c c o s x} = e^{- 2 x} \Rightarrow

k^{^{'}} = \frac{e^{- 2 x}}{\sqrt{1 - x^{2}} e^{a r c c o s x}} = \frac{e^{- 2 x - a r c c o s x}}{\sqrt{1 - x^{2}}} \Rightarrow

k (x) = \int \frac{e^{- 2 x - a r c c o s x}}{\sqrt{1 - x^{2}}} d x + c \Rightarrow

y (x) = e^{a r c o s x} {\int_{.}^{x} \frac{e^{- 2 t - a r c c o s t}}{\sqrt{1 - t^{2}}} + c}