Answer
$y(x)=C_1e^{-5x}+C_2 xe^{-5x}$
Work Step by Step
Solve the characteristic equation for the differential equation. $$r^2+10r+25=0$$
Factor and solve for the roots. $$(r+5)^2=0$$ $$r_1=-5, r_2=-5$$ as a multiplicity of $2$.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-5x}$ and $y_2=x e^{-5x}$.
Therefore, the general equation is equal to $y(x)=C_1e^{-5x}+C_2 xe^{-5x}$.