Solution of the differential equation
$(2 x - y + 2) d x + (4 x - 2 y - 1) d y = 0$ is

2 x - y = {c e}^{- (x + 2 y)}

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2 x + y = {c e}^{(2 x + y)}

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x - 2 y = {c e}^{- (x + 2 y)}

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2 x + y = {c e}^{(x + 2 y)}

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Solution

The correct option is A $2 x - y = {c e}^{- (x + 2 y)}$
Given equation is $\frac{d y}{d x} = \frac{- {(2 x - y) + 2}}{2 (2 x - y) - 1}$
Put $2 x - y = v \Rightarrow 2 - \frac{d y}{d x} = \frac{d v}{d x}$
$\Rightarrow - (\frac{d v}{d x} - 2) = - (\frac{v + 2}{2 v - 1})$
$\Rightarrow \frac{d v}{d x} = (\frac{v + 2}{2 v - 1} + 2) \Rightarrow \frac{d v}{d x} = \frac{5 v}{2 v - 1}$
$\Rightarrow \int \frac{2 v - 1}{5 v} d v = \int d x$
$\Rightarrow \frac{2}{5} v - \frac{1}{5} log v = x + log c$
$\Rightarrow \frac{2}{5} (2 x - y) - \frac{1}{5} log (2 x - y) = x + log c$
$\Rightarrow 2 (2 x - y) - log (2 x - y) = 5 x + 5 log c$
$\Rightarrow 4 x - 2 y - log (2 x - y) = 5 x - log c$
$\Rightarrow log c - log (2 x - y) = (x + 2 y)$
$\Rightarrow log (\frac{c}{2 x - y}) = (x + 2 y)$
$\Rightarrow (\frac{c}{2 x - y}) = e^{(x + 2 y)}$
$\Rightarrow c . e^{- (x + 2 y)} = (2 x - y)$

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