Show that the given differential equation is homogeneous and then solve it.

$x \frac{d y}{d x} - y + x s i n (\frac{y}{x}) = 0$

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Solution

Given, $\frac{d y}{d x} = \frac{y}{x} - s i n \frac{y}{x}$ ...(i)
Thus , the given differential equations is homogeneous
So, put $\frac{y}{x} = v \Rightarrow y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
Then, Eq. (i) becomes $v + x \frac{d v}{d x} = v - s i n v \Rightarrow c o s e c v d v = - \frac{1}{x} d x$
On integrating both sides, we get
$\int c o s e c v d v = - \int \frac{d x}{x} \Rightarrow l o g | c o s e c v - c o t v | = - l o g | x | + A$ $(∵ \int c o s e c x d x = l o g | c o s e c x - c o t x |)$
$\Rightarrow l o g | (c o s e c v - c o t v) | + l o g | x | = A \Rightarrow l o g | (c o s e c v - c o t v) x | = A \Rightarrow | x (c o s e c v - c o t v) | = e^{A} \Rightarrow x (\frac{1}{s i n v} - \frac{c o s v}{s i n v}) ≃ e^{A} \Rightarrow x (\frac{1 - c o s v}{s i n v}) = e^{A} \Rightarrow x (1 - c o s v) = e^{A} s i n v \Rightarrow x (1 - c o s (\frac{y}{x})) = C s i n (\frac{y}{x})$ $P u t C = e^{A} a n d v = \frac{v}{x}$
This is the required solutions of given differential equation.

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