what is integrating factor of (xy^2 −y) dx − x dy = 0


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Question Number 103958 by bemath last updated on 18/Jul/20

$w h a t i s i n t e g r a t i n g f a c t o r$ $o f ({x y}^{2} - y) d x - x d y = 0$
	Answered by bramlex last updated on 18/Jul/20
	$(xy^2 −y) dx = x dy (dy/dx) = ((xy^2 −y)/x) = y^2 −(y/x) (dy/dx)+(y/x) = y^2 set v = y^(−1) ⇒(dv/dx) = −y^(−2) (dy/dx) (dy/dx) = −y^2 (dv/dx) . substitute to original equation −y^2 (dv/dx) + (y/x) = y^2 (dv/dx)−(v/x) = −1 . so integrating factor is u(x) = e^(−∫ (dx/x)) = e^(ln ((1/x))) u(x) = (1/x) . solution for v(x) v(x) = ((∫ −1.((1/x))dx +C)/(1/x)) v(x) = x { ln ((1/x)) + C } (1/y) = −x ln (x) + Cx ★$
	$({x y}^{2} - y) d x = x d y$ $\frac{d y}{d x} = \frac{{x y}^{2} - y}{x} = y^{2} - \frac{y}{x}$ $\frac{d y}{d x} + \frac{y}{x} = y^{2}$ $s e t v = y^{- 1} \Rightarrow \frac{d v}{d x} = - y^{- 2} \frac{d y}{d x}$ $\frac{d y}{d x} = - y^{2} \frac{d v}{d x} . s u b s t i t u t e t o$ $o r i g i n a l e q u a t i o n$ $- y^{2} \frac{d v}{d x} + \frac{y}{x} = y^{2}$ $\frac{d v}{d x} - \frac{v}{x} = - 1 . s o i n t e g r a t i n g$ $f a c t o r i s u (x) = e^{- \int \frac{d x}{x}} = e^{\ln (\frac{1}{x})}$ $u (x) = \frac{1}{x} . s o l u t i o n f o r v (x)$ $v (x) = \frac{\int - 1 . (\frac{1}{x}) d x + C}{\frac{1}{x}}$ $v (x) = x {\ln (\frac{1}{x}) + C}$ $\frac{1}{y} = - x \ln (x) + C x ★$
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