Answer
$y(x)=C_1+C_2e^{-5x}+C_3xe^{-5x}$
Work Step by Step
Given $y'''+10y''+25y'=0$
Solve the auxiliary equation for the differential equation.
$r^3+10r^2+25r=0$
Factor and solve for the roots.
$r(r+5)^2=0$
Roots are: $r_1=0$, as a multiplicity of 1 and $r_2=-5$ as a multiplicity of 2.
This implies that there are two independent solutions to the differential equation
$y_1(x)=1$
$y_2=e^{-5x}$ and
$y_3(x)=xe^{-5x}$
Therefore, the general equation is equal to $y(x)=C_1+C_2e^{-5x}+C_3xe^{-5x}$
