Question

The solution of the differential equation $\frac{dy}{dx}+1=e^{x+y}$, is
(a) (x + y) e^{x + y} = 0
(b) (x + C) e^{x + y} = 0
(c) (x − C) e^{x + y} = 1
(d) (x − C) e^{x + y} + 1 =0

Answer 1

(d) (x − C) e^{x + y} + 1 = 0


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+1=e^{x+y} \\ \mathrm{Let}x+y=v \\ \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx} \\ \Rightarrow \frac{dy}{dx}+1=\frac{dv}{dx} \\ \therefore \frac{dv}{dx}=e^v \\ \Rightarrow e^{-v}dv=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ -e^{-v}=x-C \\ \Rightarrow -1=e^v\left(x-C\right) \\ \Rightarrow \left(x-C\right)e^{x+y}+1=0 \end{gathered}$$