For the differential equation in given question find the general solution.

$y l o g y d x - x d y = 0$

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Solution

Given, $y l o g y d x - x d y = 0$
On separating the variables, we get y~ log y~ dx=x dy $\Rightarrow \frac{1}{x} d x = \frac{1}{y l o g y} d y$
On integrating, we get $\int \frac{1}{x} d x = \int \frac{1}{y l o g y} d y L e t l o g y = t \Rightarrow \frac{1}{y} = \frac{d t}{d y} \Rightarrow\Rightarrow d y = y d t ∴ \int \frac{1}{y} d x = \int \frac{1}{t} d t \Rightarrow l o g | y | = l o g | t | - l o g | C | \Rightarrow l o g | x C | = l o g | l o g y | [∵ l o g m + l o g n = l o g m n a n d l o g m = l o g n \Rightarrow m = n] \Rightarrow C x = l o g y \Rightarrow y = e^{C x} [∵ L o g_{e} x = m \Rightarrow e^{m} = x]$
which is the required general solution.

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