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Question Number 110288 by mathdave last updated on 28/Aug/20

	Answered by bemath last updated on 28/Aug/20

	$Δ \frac{b e}{m a t h} ▽$ $l e t y = v x \Rightarrow d y = v d x + x d v$ $\Leftrightarrow (4 v x - 2 x) d x = (x + v x) (v d x + x d v)$ $(4 v - 2) d x = (1 + v) (v d x + x d v)$ $(4 v - 2) d x = (v + v^{2}) d x + (x + v x) d v$ $(3 v - 2 - v^{2}) d x = x (1 + v) d v$ $\Leftrightarrow - \frac{d x}{x} = \frac{(1 + v) d v}{v^{2} - 3 v + 2}$ $\Leftrightarrow - \ln x + c = \int \frac{(1 + v) d v}{(v - 1) (v - 2)}$ $\Leftrightarrow - \ln x + c = 3 \ln (v - 2) - 2 \ln (v - 1)$ $\ln (\frac{C}{x}) = \ln (\frac{{(v - 2)}^{3}}{{(v - 1)}^{2}})$ $\Leftrightarrow C {(v - 1)}^{2} = x {(v - 2)}^{3}$ $C {(\frac{y - x}{x})}^{2} = x {(\frac{y - 2 x}{x})}^{3}$ $C {(y - x)}^{2} = {(y - 2 x)}^{3}$ $y (1) = 3 \Leftrightarrow C {(3 - 1)}^{2} = {(3 - 2)}^{3}$ $4 C = 1 \to C = \frac{1}{4}$ $∴ {(y - x)}^{2} = 4 {(y - 2 x)}^{3}$
	Commented by mathdave last updated on 28/Aug/20

	$g o o d w o r k$
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