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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

({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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