solve-xy-y-2x-1-x-4-

Question Number 50431 by Abdo msup. last updated on 16/Dec/18

s o l v e {x y}^{^{'}} + y = \frac{2 x}{\sqrt{1 - x^{4}}}

Commented by Abdo msup. last updated on 16/Dec/18

(h e) {x y}^{^{'}} + y = 0 \Rightarrow {x y}^{^{'}} = - y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{1}{x} \Rightarrow l n ∣ y ∣= - l n ∣ x ∣ + k \Rightarrow

y (x) = \frac{K}{∣ x ∣} l e t t a k e x > 0 \Rightarrow y = \frac{K}{x} m v c m e t h o d g i v e

y^{^{'}} = \frac{K^{^{'}}}{x} - \frac{K}{x^{2}} a n d (e) \Leftrightarrow K^{^{'}} - \frac{K}{x} + \frac{K}{x} = \frac{2 x}{\sqrt{1 - x^{4}}} \Rightarrow

K^{^{'}} = \frac{2 x}{\sqrt{1 - x^{4}}} \Rightarrow K (x) = \int \frac{2 x d x}{\sqrt{1 - x^{4}}} c h a n g e m e n t x^{2} = s i n t

g i v e K (x) = \int \frac{c o s t d t}{\sqrt{1 - {s i n}^{2} t}} = \int d t + c = t + λ = a r c s i n (x^{2}) + λ

\Rightarrow y (x) = \frac{a r c s i n (x^{2}) + λ}{x} = \frac{λ}{x} + \frac{a r c s i n (x^{2})}{x} .

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Dec/18

x d y + y d x = \frac{2 x}{\sqrt{1 - {(x^{2})}^{2}}} d x

d (x y) = \frac{d (x^{2})}{\sqrt{1 - {(x^{2})}^{2}}}

\int d (x y) = \int \frac{d (x^{2})}{\sqrt{1 - {(x^{2})}^{2}}}

x y = {s i n}^{- 1} (x^{2}) + c