The particular solution of $\dfrac{dy}{dx} = e^{x+y}$ given $y(0) = 0$…

Question

The particular solution of $\dfrac{dy}{dx} = e^{x+y}$ given $y(0) = 0$ is:

Accepted Answer

D. Both B and C describe the same curve

From $e^{-y} = -e^x + c$: at $x = 0, y = 0$ ⇒ $1 = -1 + c$ ⇒ $c = 2$. So $e^{-y} + e^x = 2$, equivalent to $y = -\log(2 - e^x) = \log(1/(2-e^x))$.

Answer

$e^{-y} + e^x = 2$

Answer

$e^{-y} = 2 - e^x$ (or equivalently $e^y + e^{-x} = 2$)

Answer

$y = \log(2 - e^x)$