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Question:
$\[ \frac{dx}{dy} = \frac{x^2-3y^2}{2xy} \]$
Solution: \\
Let u be $\frac{y}{x}$. Then, $\frac{dx}{dy} = \frac{1-3u^2}{2u}$.
Also, $\frac{dy}{dx} = \frac{du}{dx} x + u$
That is, $\frac{du}{dx} x = - \frac{1-5u^2}{2u} $
$\textbf{Then} $
$\[ - \int \frac{2u}{1-5u^2} \;\mathrm{d}u= \int \frac{1}{x} \;\mathrm{d}x \]$
$\[ -\frac{1}{5} \ln (1-5u^2) = \ln x + C\]$
$\[ \ln \frac{(1-5u^2)^\frac{-1}{5}}{x} = C \]$
$\[((1-5\frac{y^2}{x^2})^\frac{1}{5}x = C\]$
