(y^2 −xy)dx + 2xy dy = 0


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Question Number 97751 by bobhans last updated on 09/Jun/20

$(y^{2} - xy) dx + 2 xy dy = 0$
Commented by bemath last updated on 09/Jun/20

$\Leftrightarrow 2 xy dy = (xy - y^{2}) dx$ $set y = zx \Leftrightarrow \frac{dy}{dx} = z + x \frac{dz}{dx}$ $\Leftrightarrow z + x \frac{d y}{d x} = \frac{x^{2} z - x^{2} z^{2}}{2 x^{2} z}$ $z + x \frac{d y}{d x} = \frac{z - z^{2}}{2 z}$ $x \frac{d y}{d x} = \frac{z - 3 z^{2}}{2 z} \Rightarrow \frac{2 z d z}{z - 3 z^{2}} = \frac{d x}{x}$ $\frac{2 d z}{3 z - 1} = - \frac{d x}{x} \Rightarrow \frac{2}{3} \ln (3 z - 1) = - \ln x + c$ $\frac{2}{3} \ln (\frac{3 y}{x} - 1) = \ln (\frac{C}{x})$ $\ln (\frac{3 y}{x} - 1) = \ln \sqrt{{(\frac{C}{x})}^{3}}$ $\frac{3 y}{x} = 1 + \sqrt{{(\frac{C}{x})}^{3}}$ $y = \frac{1}{3} (x + x \sqrt{{(\frac{C}{x})}^{3}})$
Commented by bobhans last updated on 09/Jun/20

$waw . . . . . . .$
Commented by john santu last updated on 10/Jun/20

$yes . . correct$
	Answered by smridha last updated on 09/Jun/20

	$\frac{\partial}{\partial y} (y^{2} - x y) = 2 y - x \neq \frac{\partial}{\partial x} (2 x y) = 2 y$ $t h i s i s n o t e x a c t d i f f e r e n t i a l$ $s o, \frac{1}{2 x y} [2 y - x - 2 y] = - \frac{1}{2 y}$ $I . F = e^{\int - \frac{d y}{2 y}} = \frac{1}{\sqrt{y}} m u l t i : b y t h i s o f b o t h$ $s i d e s o f g i v e n {e q}^{n}$ $(y^{\frac{3}{2}} - {x y}^{\frac{1}{2}}) d x + 2 {x y}^{\frac{1}{2}} d y = 0$ $n o w \int_{(k e e p i n g y a s c o n s t a n t)} (y^{\frac{3}{2}} - {x y}^{\frac{1}{2}}) d x$ $= {x y}^{\frac{3}{2}} - \frac{x^{2}}{2} y^{\frac{1}{2}}$ $a n d \int_{(t e r m s a r e d o n o t c o n t a i n i n g x)} (0) d y = 0$ $s o s o l u t i o n :$ $x y^{\frac{3}{2}} - \frac{x^{2}}{2} y^{\frac{1}{2}} = c$
	Commented by mr W last updated on 09/Jun/20

	$p l e a s e c h e c k s i r :$ $w i t h {x y}^{\frac{3}{2}} - \frac{x^{2}}{2} y^{\frac{1}{2}} = c w e g e t :$ $y^{\frac{3}{2}} + \frac{3}{2} {x y}^{\frac{1}{2}} \frac{d y}{d x} - {x y}^{\frac{1}{2}} - \frac{x^{2}}{4} y^{- \frac{1}{2}} \frac{d y}{d x} = 0$ $4 y (y - x) + x (6 y - x) \frac{d y}{d x} = 0$ $\Rightarrow 4 y (y - x) d x + x (6 y - x) d y = 0$ $t h i s i s n o t t h e o r i g i n a l d . e q n .$
	Commented by smridha last updated on 09/Jun/20

	$I d o n o t f i n d a n y m i s t a k e . .$ $i f y o u c a n t h e n s h o w m e . . . .$
	Commented by prakash jain last updated on 10/Jun/20

	$(y^{2} - x y) d x + 2 x y d y = 0$ $M d x + N d y = 0$ $M_{y} = 2 y - x, N_{x} = 2 y$ $\frac{M_{y} - N_{x}}{N} = \frac{- x}{2 x y} = - \frac{1}{2 y}$ $I think the mistake is in$ $calculating I . F .$ $It is function of y . It should be$ $function of x to use as integrating$ $factor .$ $(y^{\frac{3}{2}} - {x y}^{\frac{1}{2}}) d x + 2 {x y}^{\frac{1}{2}} d y = 0$ $So the above equation is still not exact .$
	Commented by smridha last updated on 10/Jun/20

	$y e a h m y s e c o n d s o l u t i o n t e l l s$ $t h e t r u t h . . . . b u t i a m n o t y e t$ $u n d e r s t a n d w h y t h e I F i s d i f f e r e n t$ $f o r t h i s t w o m e h o d s . . . . . t w o$ $m e t h o d s s h o u l d b e a n a l o g u s$ $t o e a c h o t h e r . . . .$
	Commented by prakash jain last updated on 10/Jun/20

	$\frac{M_{y} - N_{x}}{N} should not have y and$ $should be a fuction of only x .$ $So there is a mistake in method$ $used for computing I . F in the answer$ $above$
	Answered by smridha last updated on 10/Jun/20

	$(y^{2} - x y) d x + 2 x y d y = 0$ $\frac{d y}{d x} + \frac{y}{2 x} = \frac{1}{2} s o I F = e^{\int \frac{d x}{2 x}} = \sqrt{x}$ $m u l t : a n d i n t e g r a t i n g b y b o t h s i d e s w i t h t h i s w e g e t$ $\int d (y \sqrt{x}) = \frac{1}{2} \int \sqrt{x} d x$ $s o t h e s o l u t i o n i s$ $y \sqrt{x} = \frac{1}{3} x^{\frac{3}{2}} + c$ $n o w t h i s i s p e r f e c t . . . . b u t s t i l l I$ $h a v e c o n f u t i o n w h y m y p r e v i o u s$ $m e t h o d d i d n o t w o r k . . . p r o b a b l y$ $I d i d s i l l y m i s t a k e a n y w h e r e . . . .$
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