If the solution of $x^{2} y^{' 2} + 2 y^{2} = x^{2} + 2 x y y^{'}$ is $y = x sin (ln k x^{n})$ then $n$ is equal to ( $n \in N$ ):

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Solution

$x^{2} y^{' 2} + 2 y^{2} = x^{2} + 2 x y y^{'} y$
$y^{'} = \frac{y \pm \sqrt{x^{2} - y^{2}}}{x}$
Substitute $y = v x$ to get $s i n^{- 1} v = ln x + ln k$
$\Rightarrow v = s i n (ln k x) \Rightarrow y = x s i n (ln (k x))$
$\Rightarrow n = 1$

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