y-bx-ay-ax-by-

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

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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