# Math Help - ODE

1. ## ODE

hello
I have big problem with this one:

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

2. Mathematica says the solution is

$y = x - \ln{\left(-C - e^{e^x}\right)}$.

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

multiply both sides by $e^{-y}$

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

Substitute $u=e^{-y}$, to get:

$\frac{du}{dx}=-u-e^{e^x}$

$\frac{du}{dx}+u=-e^{e^x}$

Find your integrating factor and continue . . .

4. Thanks..