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Integrating Factor

The solution ...

Question

The solution of an differential equation (1 + xy)x dy + (1 - xy)y dx = 0, is

\frac{1}{x y}

+ log

\frac{y}{x}

= C

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- x y + l o g \frac{y}{x} = C

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\frac{- 1}{x y}

+ log

\frac{y}{x}

= C

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\frac{- 1}{x y}

+ log

\frac{x}{y}

= C

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Solution

The correct option is C $\frac{- 1}{x y}$ + log $\frac{y}{x}$ = C
Arrange given differential equation as-

$\Rightarrow x d y + y d x + x^{2} y d y - x y^{2} d x$ = 0

$\Rightarrow (x y) + x y (x d y - y d x)$ = 0

$\Rightarrow \frac{d (x y)}{x y} + x^{2} d (\frac{y}{x})$ =0

Divide this equation by $x y$

$\Rightarrow \frac{d (x y)}{(x y)^{2}} + \frac{x}{y} d (\frac{y}{x})$ =0

$\Rightarrow \frac{d (x y)}{(x y)^{2}} + \frac{1}{\frac{y}{x}} d (\frac{y}{x})$ =0

Integrate this equation-

$\Rightarrow \frac{- 1}{x y} + l o g \frac{y}{x}$ =0

Suggest Corrections

Similar questions

y (x y + 2 x^{2} y^{2}) d x + x (x y - x^{2} y^{2}) d y = 0

p . | l o g | x | + Q . l o g | y | - \frac{1}{x y} = C

x y = log y + c

(x y - 1) \frac{d y}{d x} + y^{2} = 0

\frac{d y}{d x} = \frac{x^{2} + x y + y^{2}}{x^{2}}

\tan^{- 1} (\frac{x}{y}) = \log y + C

\tan^{- 1} (\frac{y}{x}) = \log x + C

\tan^{- 1} (\frac{x}{y}) = \log x + C

\tan^{- 1} (\frac{y}{x}) = \log y + C

(1 + x) y d x + (1 + y) x d y = 0

log x y + (x - y) = c .

x y = l o g

y + c

\frac{d y}{d x} = \frac{y^{2}}{1 - x y} (x y \neq 1)

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