Question

The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is

Options

• (x + y) e^x + y = 0

• (x + C) e^x + y = 0

• (x − C) e^x + y = 1

• (x − C) e^x + y + 1 =0

Answer 1

(x − C) e^x + 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
\[ - e^{- v} = x - C\]
\[ \Rightarrow - 1 = e^v \left( x - C \right)\]
\[ \Rightarrow \left( x - C \right) e^{x + y} + 1 = 0\]