Answer
$\frac{1-\sqrt{5}}{2}$, $\frac{1+\sqrt{5}}{2}$
Work Step by Step
$y=e^{\lambda x}$
$y'=e^{\lambda x}\frac{d}{dx}(\lambda x)$
$y'=\lambda e^{\lambda x}$
$y''=\lambda e^{\lambda x}\frac{d}{dx}(\lambda x)$
$y''=\lambda^2 e^{\lambda x}$
$y+y'=y''$
$e^{\lambda x}+\lambda e^{\lambda x}=\lambda^2 e^{\lambda x}$
$1+\lambda=\lambda^2$
$\lambda^2-\lambda-1=0$
$\lambda=\frac{-(-1)\pm\sqrt{(-1)^2-4*1*(-1)}}{2*1}$
$\lambda=\frac{1\pm\sqrt{1+4}}{2}$
$\lambda=\frac{1\pm\sqrt{5}}{2}$
So the solutions are $\lambda=\frac{1-\sqrt{5}}{2}$ and $\lambda=\frac{1+\sqrt{5}}{2}$.
