y′ = xy^2 −(y/x)


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Question Number 120905 by bemath last updated on 03/Nov/20

$y^{'} = {xy}^{2} - \frac{y}{x}$
	Answered by Bird last updated on 03/Nov/20

	$\Rightarrow \frac{y^{^{'}}}{y^{2}} = x - \frac{1}{y x} l e t \frac{1}{y} = z \Rightarrow z^{^{'}} = - \frac{y^{^{'}}}{y^{2}}$ $e \Rightarrow - z^{^{'}} = x - \frac{z}{x} \Rightarrow z^{^{'}} = - x + \frac{z}{x}$ $\Rightarrow z^{^{'}} - \frac{z}{x} = - x \Rightarrow {x z}^{^{'}} - z = - x^{2}$ $h \to {x z}^{^{'}} - z = 0 \Rightarrow {x z}^{^{'}} = z \Rightarrow \frac{z^{^{'}}}{z} = \frac{1}{x}$ $\Rightarrow l n ∣ z ∣= l n ∣ x ∣ + c \Rightarrow z = k x$ $m v c m e y h o d \Rightarrow z^{^{'}} = k^{^{'}} x + k$ $e \to k^{^{'}} x^{2} + k x - k x = - x^{2} \Rightarrow$ $k^{^{'}} = - 1 \Rightarrow k = - x + λ \Rightarrow$ $z = (- x + λ) x = - x^{2} + λ x \Rightarrow$ $y = \frac{1}{- x^{2} + λ x}$
	Answered by bobhans last updated on 04/Nov/20

	$y^{'} + \frac{y}{x} = {x y}^{2}$ $l e t y^{- 1} = u \Rightarrow \frac{d u}{d x} = - y^{- 2} \frac{d y}{d x}$ $o r \frac{d y}{d x} = - y^{2} \frac{d u}{d x}$ $s u b s t i t u t i n g t o d i f f e q u a t i o n$ $g i v e s - y^{2} \frac{d u}{d x} + \frac{y}{x} = {x y}^{2}$ $\Leftrightarrow \frac{d u}{d x} - \frac{u}{x} = - x, p u t i n t e g r a t i n g$ $f a c t o r v = e^{\int - \frac{d x}{x}} = e^{\ln (\frac{1}{x})} = \frac{1}{x}$ $m u l t i p l y b y \frac{1}{x} b o t h s i d e s$ $\Leftrightarrow \frac{1}{x} \frac{d u}{d x} - \frac{1}{x^{2}} u = - 1$ $\Leftrightarrow \frac{d}{d x} (\frac{u}{x}) = - 1 . I n t e g r a t i n g b o t h$ $s i d e s g i v e s \int d (\frac{u}{x}) = - \int d x$ $\Leftrightarrow \frac{u}{x} = - x + c \to u = - x^{2} + c x$ $\Leftrightarrow \frac{1}{y} = - x^{2} + c x, t h u s y = \frac{1}{- x^{2} + c x} .$
	Answered by Olaf last updated on 04/Nov/20

	$y^{'} = {x y}^{2} - \frac{y}{x}$ ${x y}^{'} + y = {(x y)}^{2}$ $\frac{{(x y)}^{'}}{{(x y)}^{2}} = 1$ $- \frac{1}{x y} = x + C$ $y = - \frac{1}{x (x + C)}$
	Answered by TANMAY PANACEA last updated on 04/Nov/20

	$\frac{d y}{d x} + \frac{y}{x} = {x y}^{2}$ $\frac{x d y + y d x}{x^{2} y^{2}} = d x$ $\int \frac{d (x y)}{(x y)} = \int d x$ $\frac{- 1}{x y} = x + c$
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