Find the solution of $(2 x + y + 1) d x + (4 x + 2 y - 1) d y = 0$ .

2 y - x + log (2 x + y - 1) = k

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y + x + log (2 x + y - 1) = k

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y - x + log (2 x + y - 1) = k

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 y + x + log (2 x + y - 1) = k

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Solution

The correct option is D $2 y + x + log (2 x + y - 1) = k$
Given $(2 x + y + 1) d x + (4 x + 2 y - 1) d y = 0$
$\frac{d y}{d x} = \frac{- (2 x + y + 1)}{2 (2 x + y) - 1}$ .....(1)
Put $2 x + y = v$
$2 + \frac{d y}{d x} = \frac{d v}{d x}$
So, eqn (1) becomes
$\frac{d v}{d x} - 2 = \frac{- v - 1}{2 v - 1}$
$\Rightarrow \frac{d v}{d x} = \frac{3 v - 3}{2 v - 1}$
$\Rightarrow \frac{2 v - 1}{3 v - 3} d v = d x$
$\Rightarrow \frac{2}{3} \frac{v - \frac{1}{2}}{v - 1} d v = d x$
$\Rightarrow \frac{v - \frac{1}{2}}{v - 1} d v = \frac{3}{2} d x$
$\Rightarrow \frac{v - 1 + \frac{1}{2}}{v - 1} d v = \frac{3}{2} d x$
Integrating , we get
$v + \frac{1}{2} log (v - 1) = \frac{3}{2} x$
$\Rightarrow 2 x + y + \frac{1}{2} log (2 x + y - 1) = \frac{3}{2} x$
$2 y + x + log (2 x + y - 1) = k$

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