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

$y d x + x l o g (\frac{y}{x}) d y - 2 x d y = 0$

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Solution

Given, $y d x + x l o g (\frac{y}{x}) d y - 2 x d y = 0$
$\Rightarrow \frac{y}{x} + l o g (\frac{y}{x}) \frac{d y}{d x} - 2 \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{\frac{y}{x}}{2 - l o g \frac{y}{x}}$ ...(i)
Thus, the given differential equation is homogeneous.
So, put $\frac{y}{x} = v i . e ., y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
Then, Eq.(i) becomes $v + x \frac{d v}{d x} = \frac{v}{2 - l o g v} \Rightarrow x \frac{d v}{d x} = \frac{v}{2 - l o g v} - v \Rightarrow x \frac{d v}{d x} = \frac{v - 2 v + v l o g v}{2 - l o g v} \Rightarrow x \frac{d v}{d x} = \frac{v l o g v - v}{2 - l o g v} \Rightarrow \frac{2 - l o g v}{v l o g v - v} d v = \frac{1}{x} d x \Rightarrow \frac{1 - (l o g v - 1)}{v (l o g v - 1)} d v = \frac{1}{x} d x \Rightarrow (\frac{1}{v (l o g v - 1)} - \frac{(l o g v - 1)}{v (l o g v - 1)}) d v = \frac{1}{x} d x$
On integrating both sides, we get
$\int (\frac{1}{v (l o g v - 1)} - \frac{1}{v}) d v = \int \frac{d x}{x} \Rightarrow \int \frac{1}{v (l o g v - 1)} d v - \int \frac{1}{v} d v = \int \frac{d x}{x}$
Let logv-1=t $\Rightarrow \frac{1}{v} d v = d t$
$∴ \int \frac{d t}{t} - \int \frac{1}{v} d v = \int \frac{d x}{x})$
$\Rightarrow l o g | t | - l o g | v | = l o g | x | + l o g C \Rightarrow l o g | l o g v - 1 | - l o g | v | = l o g | x | + l o g C$ $[∵ t = l o g v - 1]$
$\Rightarrow l o g | l o g v - 1 | - l o g | v | - l o g | x | = l o g C \Rightarrow l o g ∣ ∣ \frac{l o g v - 1}{v x} ∣ ∣ = l o g C$ $[∵ l o g \frac{m}{n} = l o g m - l o g n]$
$\Rightarrow ∣ ∣ \frac{l o g v - 1}{v x} ∣ ∣ = C \Rightarrow \frac{l o g v - 1}{v x} = C \Rightarrow \frac{l o g (\frac{y}{x}) - 1}{y} = C$ $[∵ v = y / x]$
$\Rightarrow l o g (\frac{y}{x}) - 1 = C y$
This is the required solution of the given differential equation.

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