solve the first order differential equation: xdy − ydx = (xy)^(1/2) dx


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Question Number 206449 by necx122 last updated on 15/Apr/24

$s o l v e t h e f i r s t o r d e r d i f f e r e n t i a l$ $e q u a t i o n :$ $x d y - y d x = {(x y)}^{1 / 2} d x$
	Answered by Berbere last updated on 17/Apr/24

	$y^{'} x - y = \sqrt{x y}; y^{'} = \frac{d y}{d x}$ $\frac{y^{'} x - y}{x^{2}} = \sqrt{\frac{y}{x}} . \frac{1}{x}$ $d (\frac{y}{x}) = {(\frac{y}{x})}^{\frac{1}{2}} . \frac{1}{x} d x$ $\frac{d (\frac{y}{x})}{\sqrt{\frac{y}{x}}} = \frac{1}{x} \Rightarrow 2 \sqrt{\frac{y}{x}} = l n (x) + c \Rightarrow \frac{y}{x} = {(\frac{l n (x) + c}{2})}^{2}$ $y = x {(c^{'} + l n (\sqrt{x}))}^{2}$
	Commented by necx122 last updated on 17/Apr/24

	$p l e a s e, c s n y o u e x p l a i n t h e 3 r d l i n e ?$ $d i d y o u d o d y / d x o r d (y / x) ?$
	Commented by Berbere last updated on 17/Apr/24

	$\frac{y^{'} . x - 1 . y}{x^{2}} = {(\frac{y}{x})}^{'} = \frac{d}{d x} (\frac{y}{x})$
	Commented by necx122 last updated on 17/Apr/24
	Thank you sir. It's clear now.
	Commented by Berbere last updated on 18/Apr/24

	$w i t h e P l e a s u r$
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