The solution of the differential equation $\frac{dy}{dx} + 1 = e^{x+y} $, is
Answer & explanation
Correct answer: option 4
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, $1+\frac{dy}{dx} =\frac{dv}{dx}$
$∴\frac{dy}{dx} + 1=e^{x+y}$
$⇒\frac{dv}{dx} = e^v$
$⇒e^{-v}dv= dx$
$⇒∫e^{-v}dv=∫dx$
$⇒-e^{-v}= x+C$
$⇒-e^{-x(+y)}=x+C$
$⇒(x+C)e^{x+y}+1=0$