The solution of the differential equation $x^{2} (x d y + y d x) = (x y - 1)^{2} d x$ is (where c is an arbitrary constant):

x y - 1 = c x

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x y - 1 = c x^{2}

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\frac{1}{x y - 1} = \frac{1}{x} + c

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\frac{1}{(x y - 1)^{3}} = \frac{1}{x^{3}} + c

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Solution

The correct option is C $\frac{1}{x y - 1} = \frac{1}{x} + c$
We have,
$x^{2} (x d y + y d x) = {(x y - 1)}^{2} d x$
$x^{2} d (x y) = {(x y - 1)}^{2} d x$
$\frac{d (x y)}{{(x y - 1)}^{2}} = \frac{d x}{x^{2}}$
$\Rightarrow \int \frac{d (x y)}{{(x y - 1)}^{2}} = \int \frac{d x}{x^{2}}$ ......................(now integrate both sides , we get)
$\Rightarrow \frac{- 1}{(x y - 1)} = \frac{- 1}{x} + c$
$\Rightarrow \frac{1}{(x y - 1)} = \frac{1}{x} + k$
Hence, option C is correct

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