Answer
$y=c_{1}e^{-2x}+c_{2}xe^{-2x}$
Work Step by Step
$y''+4y'+4y=0$
Change to auxiliary equation by replacing nth derivative of $y$ with $r^{n}$.
$r^{2}+4r+4=0$
$r^{2}+4r+4=0$
$(r+2)^{2}=0$
$r=-2$
By Theorem 4, if the auxiliary equation has only one real root $r$, then the general solution is
$y=c_{1}e^{rx}+c_{2}xe^{rx}$
$y=c_{1}e^{-2x}+c_{2}xe^{-2x}$
