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Question
The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is
Options
(x + y) ex + y = 0
(x + C) ex + y = 0
(x − C) ex + y = 1
(x − C) ex + y + 1 =0
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Solution
(x − C) ex + y + 1 = 0
We have,
\[\frac{dy}{dx} + 1 = e^{x + y} \]
\[\text{ Let }x + y = v\]
\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]
\[ \Rightarrow \frac{dy}{dx} + 1 = \frac{dv}{dx}\]
\[ \therefore \frac{dv}{dx} = e^v \]
\[ \Rightarrow e^{- v} dv = dx\]
Integrating both sides, we get
\[ - e^{- v} = x - C\]
\[ \Rightarrow - 1 = e^v \left( x - C \right)\]
\[ \Rightarrow \left( x - C \right) e^{x + y} + 1 = 0\]
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