Answer
$y(x) = C_1e^{-2x}+ C_2xe^{-2x} $
Work Step by Step
We assume a solution of the form $y=e^{mx}$,where $m$ Is a constant.
$y = e^{mx}$ $\implies$ $y' = m(e^{mx})$ and $y'' = m^2(e^{mx})$
$y'' +4y' + 4y = 0$
$\implies$ $m^2(e^{mx}) + 4m(e^{mx}) + 4(e^{mx}) = 0 $
$\implies$ $e^{mx}(m^2+4m+4) = 0$
since $e^{mx}$ cant be 0
$\implies$ $m^2 +4m +4 = 0$
$\implies$ $m^2 + 2m + 2m + 4 = 0$
$\implies$ $(m+2)(m+2) = 0$
$\implies$ $(m+2)2 = 0$
$\implies$ $ m = -2$(twice)
since we have equal roots,we have that :
$y(x) = C_1e^{-2x} + c_2xe^{-2x}$
