## Calculus 8th Edition

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