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Question
\[\frac{dy}{dx} + 1 = e^{x + y}\]
Solution
We have,
\[\frac{dy}{dx} + 1 = e^{x + y} . . . . . \left( 1 \right)\]
Let `x + y = v`
\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]
Then, (1) becomes
\[\frac{dv}{dx} - 1 + 1 = e^v \]
\[ \Rightarrow \frac{dv}{dx} = e^v \]
\[ \Rightarrow e^{- v} dv = dx\]
Integrating both sides, we get
\[\int e^{- v} dv = \int dx\]
\[ \Rightarrow - e^{- v} = x + C\]
\[ \Rightarrow - 1 = e^v \left( x + C \right)\]
\[ \Rightarrow - 1 = \left( x + C \right) e^{x + y}\]
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