The general solution of the differential equation
$(x + y + 3) \frac{d y}{d x} = 1$ is

x + y + 3 = C e^{y}

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x + y + 4 = C e^{y}

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x + y + 3 = C e^{- y}

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x + y + 4 = C e^{- y}

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x + y + 4 e^{y} = C

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Solution

The correct option is C $x + y + 4 = C e^{y}$
We have $(x + y + 3) \frac{d y}{d x} = 1$
$\Rightarrow (x + y + 3) = \frac{d x}{d y} . . . (i)$
Let $x + y + 3 = t$
On differentiating both sides
$\frac{d x}{d y} + 1 = \frac{d t}{d y} \Rightarrow \frac{d t}{d y} = t + 1$
On integrating both sides
$\int \frac{d t}{t + 1} = \int d y \Rightarrow log (t + 1) = y + C_{1}$
$\Rightarrow log (x + y + 3 + 1) = y + C_{1}$
$∴ x + y + 4 = C e^{y}$

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