Resolve y=xy′+a(√(1+(y′)^2 ))


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 168742 by LEKOUMA last updated on 16/Apr/22

$R e s o l v e$ $y = {x y}^{'} + a \sqrt{1 + {(y^{'})}^{2}}$
	Answered by mindispower last updated on 19/Apr/22
	$⇒ y′=xy′′+y′+a((2y′y′′)/(2(√(1+y′^2 )))) y′=z xz′+((zz′)/( (√(1+z^2 ))))=0 z′(x+a(z/( (√(1+z^2 )))))=0 by continuity⇔ { ((z′=0)),(((z/( (√(1+z^2 ))))=−(1/a)x)) :} z=c or z^2 =x^2 (1+z^2 )⇒z^2 =((((x/a))^2 )/(1−((x/a))^2 )),x∈]−1,1[ z=((−x)/( (√(a^2 −x^2 )))),⇒y=(√(a^2 −x^2 ))+c,a>0 y=((−x^2 )/( (√(a^2 −x^2 ))))+a(√(1+(x^2 /(a^2 −x^2 ))))⇒c=0 cx+d=x(c)+a(√(1+c^2 )) d=a(√(1+c^2 )) S={cx+a(√(1+c^2 )),(√(a^2 −x^2 ))}$
	$\Rightarrow$ $y^{'} = {x y}^{″} + y^{'} + a \frac{2 y^{'} y^{″}}{2 \sqrt{1 + y^{' 2}}}$ $y^{'} = z$ ${x z}^{'} + \frac{{z z}^{'}}{\sqrt{1 + z^{2}}} = 0$ $z^{'} (x + a \frac{z}{\sqrt{1 + z^{2}}}) = 0$ $b y c o n t i n u i t y \Leftrightarrow {\begin{cases} z^{'} = 0 \\ \frac{z}{\sqrt{1 + z^{2}}} = - \frac{1}{a} x \end{cases}$ $z = c$ $o r z^{2} = x^{2} (1 + z^{2}) \Rightarrow z^{2} = \frac{{(\frac{x}{a})}^{2}}{1 - {(\frac{x}{a})}^{2}}, x \in] - 1, 1 [$ $z = \frac{- x}{\sqrt{a^{2} - x^{2}}}, \Rightarrow y = \sqrt{a^{2} - x^{2}} + c, a > 0$ $y = \frac{- x^{2}}{\sqrt{a^{2} - x^{2}}} + a \sqrt{1 + \frac{x^{2}}{a^{2} - x^{2}}} \Rightarrow c = 0$ $c x + d = x (c) + a \sqrt{1 + c^{2}}$ $d = a \sqrt{1 + c^{2}}$ $S = {c x + a \sqrt{1 + c^{2}}, \sqrt{a^{2} - x^{2}}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum