Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x + y}}$

Answer

$\frac{\text{dy}}{\text{dx}}+1 = \text{e}^\text{x+y} .....(1)$
Let $\text{ x}+\text{y} = \text{t}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}} = \frac{\text{dt}}{\text{dx}}$
Substituting the value of $\text{x + y = t}$ and $1 + \frac{\text{dy}}{\text{dx}} = \frac{\text{dt}}{\text{dx}} (1),$ we get
$\frac{\text{dt}}{\text{dx}} = \text{e}^1$
$\Rightarrow \text{e}^{-1}\text{dt} = \text{dx}$
$\Rightarrow -\text{e}^{-1} = \text{x}+\text{C}$
$\Rightarrow -\text{e}^{-(\text{x+y})} = \text{x} +\text{C}$ $[\therefore \text{t} = \text{x} + \text{y}]$

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