The solution of

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Solution

A) The equation can be written as
$x^{4} e^{x} d x + 4 x y (x d y - y d x) = 0$
Dividing by $x^{4},$ we have $e^{x} d x + 4 \frac{y}{x} . \frac{x d y - y d x}{x^{2}} = 0$
$e^{x} + 2 {(\frac{y}{x})}^{2} = C$
B) The equation can written as
$y d x + x d y + x y (y d x - x d y) = 0$
$d (x y) + x^{2} y^{2} (\frac{d x}{x} - \frac{d y}{y}) = 0 \Rightarrow log \frac{x}{y} - \frac{1}{x y} = C$
C) Putting $3 x - 2 y = u$ in equation (3), we have
$\frac{1}{2} [3 - \frac{d u}{d x}] = \frac{2 u + 3}{u + 1} \Rightarrow \frac{d u}{d x} = 2 [\frac{3}{2} - \frac{2 u + 3}{u + 1}] = - \frac{u + 3}{u + 1}$
$\frac{u + 3}{u + 1} d u = - d x \Rightarrow u - 2 log (u + 3) = - x + C$
$4 x - 2 y - 2 log (3 x - 2 y + 3) = C$

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