y′ = ((bx+ay)/(ax+by))


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Question Number 133631 by liberty last updated on 23/Feb/21

$y^{'} = \frac{bx + ay}{ax + by}$
	Answered by TheSupreme last updated on 23/Feb/21

	$c a s o 1 : b = 0$ $y^{'} = \frac{a y}{a x} = \frac{y}{x}$ $l n (y) = l n (x) + c$ $y = c_{1} x$ $c a s o 2 : b \neq 0$ $ξ = \frac{a}{b}$ $y^{'} = \frac{x + ξ y}{ξ x + y}$ $ξ x + y = u$ $y^{'} = u^{'} + ξ$ $u^{'} + ξ = \frac{x + ξ (u + ξ x)}{u}$ $u^{'} + ξ = \frac{1}{u} (x + ξ u + ξ^{2} x)$ ${u u}^{'} + ξ u = x + ξ u + ξ^{2} x$ ${u u}^{'} = x (1 + ξ^{2})$ $\frac{u^{2}}{2} = \frac{x^{2}}{2} (1 + ξ^{2}) + c$ $u^{2} = x^{2} (1 + ξ^{2}) + c$ $u =∣ x ∣ \sqrt{c + 1 + ξ^{2}}$ $y =∣ x ∣ \sqrt{c + 1 + ξ^{2}} + ξ x + c$ $v e r i f y f o r c = 0$ $y^{'} = s g n (x) \sqrt{1 + ξ^{2}} + ξ$ $s g n (x) \sqrt{1 + ξ^{2}} + ξ = \frac{x + ξ ∣ x ∣ \sqrt{1 + ξ^{2}} + ξ^{2} x}{∣ x ∣ \sqrt{1 + ξ^{2}}}$ $= \frac{x [(1 + ξ^{2}) + ξ s g n (x) \sqrt{1 + ξ^{2}}]}{∣ x ∣ \sqrt{1 + ξ^{2}}} = s g n (x) \sqrt{1 + ξ^{2}} + ξ$
	Answered by mr W last updated on 23/Feb/21

	$y = x u$ $y^{'} = u + x \frac{d u}{d x} = \frac{b + a u}{a + b u}$ $x \frac{d u}{d x} = \frac{b (1 - u^{2})}{a + b u}$ $\frac{(a + b u) d u}{1 - u^{2}} = \frac{b d x}{x}$ $\int \frac{(a + b u) d u}{1 - u^{2}} = \int \frac{b d x}{x}$ $\int (\frac{a}{1 - u^{2}} + \frac{b u d u}{1 - u^{2}}) d u = \int \frac{b d x}{x}$ $\frac{a}{2} \int (\frac{1}{1 - u} + \frac{1}{1 + u}) d u - \frac{b}{2} \int \frac{d (1 - u^{2})}{1 - u^{2}} = \int \frac{b d x}{x}$ $\frac{a}{2} \ln ∣ \frac{1 + u}{1 - u} ∣ - \frac{b}{2} \ln ∣ 1 - u^{2} ∣= b \ln x + C$ $a \ln ∣ \frac{1 + u}{1 - u} ∣ - b \ln ∣ 1 - u^{2} ∣= 2 b \ln x + C$ $\ln \frac{{(1 + u)}^{a - b}}{{(1 - u)}^{a + b}} = \ln {c x}^{2 b}$ $\Rightarrow \frac{{(1 + u)}^{a - b}}{{(1 - u)}^{a + b}} = {c x}^{2 b}$ $\Rightarrow \frac{{(1 + \frac{y}{x})}^{a - b}}{{(1 - \frac{y}{x})}^{a + b}} = {c x}^{2 b}$ $\Rightarrow \frac{{(x + y)}^{a - b}}{{(x - y)}^{a + b}} = c$
	Answered by bobhans last updated on 23/Feb/21
	$let y = ux ⇒(dy/dx) = u + x (du/dx) ⇔u+x (du/dx) = ((bx+aux)/(ax+bux))=((b+au)/(a+bu)) ⇔ x (du/dx) = ((b+au−au−bu^2 )/(a+bu)) ⇔ x (du/dx) = ((b−bu^2 )/(a+bu)) ⇒ ((a+bu)/(b(1−u^2 ))) du = (dx/x) Partial fraction ((a+bu)/(1−u^2 )) = (p/(1+u)) + (q/(1−u)) p = [((a+bu)/(1−u)) ]_(u=−1) = ((b−a)/2) q= [ ((a+bu)/(1+u)) ]_(u=1) = ((b+a)/2) ⇒(1/b)∫(((b−a)/(2(1+u))) + ((b+a)/(2(1−u))))du=∫ (dx/x) ((b−a)/2) ln ∣1+u∣−((b+a)/2) ln ∣1+u∣= b ln ∣Cx∣ (b−a)ln ∣((y+x)/x)∣ −(b+a)ln ∣((y−x)/x)∣=2b ln ∣Cx∣$
	$l e t y = u x \Rightarrow \frac{d y}{d x} = u + x \frac{d u}{d x}$ $\Leftrightarrow u + x \frac{d u}{d x} = \frac{b x + a u x}{a x + b u x} = \frac{b + a u}{a + b u}$ $\Leftrightarrow x \frac{d u}{d x} = \frac{b + a u - a u - {b u}^{2}}{a + b u}$ $\Leftrightarrow x \frac{d u}{d x} = \frac{b - {b u}^{2}}{a + b u}$ $\Rightarrow \frac{a + b u}{b (1 - u^{2})} d u = \frac{d x}{x}$ $P a r t i a l f r a c t i o n$ $\frac{a + b u}{1 - u^{2}} = \frac{p}{1 + u} + \frac{q}{1 - u}$ $p = {[\frac{a + b u}{1 - u}]}_{u = - 1} = \frac{b - a}{2}$ $q = {[\frac{a + b u}{1 + u}]}_{u = 1} = \frac{b + a}{2}$ $\Rightarrow \frac{1}{b} \int (\frac{b - a}{2 (1 + u)} + \frac{b + a}{2 (1 - u)}) d u = \int \frac{d x}{x}$ $\frac{b - a}{2} \ln ∣ 1 + u ∣ - \frac{b + a}{2} \ln ∣ 1 + u ∣= b \ln ∣ C x ∣$ $(b - a) \ln ∣ \frac{y + x}{x} ∣ - (b + a) \ln ∣ \frac{y - x}{x} ∣= 2 b \ln ∣ C x ∣$
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