Question

Solve the differential equation $\frac{dy}{dx}+1=e^{x+y}$

Answer 1

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

Take $x+y=t$

$1+\frac{dy}{dx}=\frac{dt}{dx}$

Therefore, in equation 1, we get,

$\frac{dt}{dx}=e^t$

$e^{-t}dt=dx$

$-e^{-t}=x+C$

$\frac{-1}{e^{x+y}}=x+C$

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

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