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

$(x-C)e^{x+y}+1=0$

The correct answer is option (4) : $(x-C)e^{x+y}+1=0$

We have,

$\frac{dy}{dx}+ 1= e^{x+y}$

Let $ x+y = v.$ Then, $1+\frac{dy}{dx} =\frac{dv}{dx}$

$∴\frac{dy}{dx} + 1=e^{x+y}$

$⇒\frac{dv}{dx} = e^v$

$⇒e^{-v}dv= dx$

$⇒∫e^{-v}dv=∫dx$

$⇒-e^{-v}= x+C$

$⇒-e^{-x(+y)}=x+C$

$⇒(x+C)e^{x+y}+1=0$