Question

The solution of $\left(\frac{dy}{dx}\right)+1=e^{x+y}$is

• A. $e^{-(x+y)}+x+c=0$
• B. $e^{-(x+y)}–x+c=0$
• C. $e^{x+y}+x+c=0$
• D. $e^{x+y}–x+c=0$

Answer 1

The correct option is A

$e^{-(x+y)}+x+c=0$

Explanation for the correct option

\Find the solution of the given differential equation

Given:$\left(\frac{dy}{dx}\right)+1=e^{x+y}$ ….. (i)

Let, $x+y=Z$

$\therefore \frac{dy}{dx}+1=\frac{dZ}{dx}$ …… (ii)

Now, comparing (i) and (ii), we get,

$$\begin{gathered} \frac{dZ}{dx}=e^Z \\ \Rightarrow dx=e^{-Z}dZ \end{gathered}$$

On integrating both sides, we get

$$\begin{gathered} \int dx=\int e^{-Z}dZ \\ \Rightarrow x+c=-e^{-Z} \\ \Rightarrow x+e^{-\left(x+y\right)}+c=0 \end{gathered}$$

Hence, option (A) is the correct answer.