Answer
$y(x)=C_1e^{-5x}+C_2xe^{-5x}+C_3x^2 e^{-5x}+C_4e^{-3x}+C_5e^x$
Work Step by Step
Solve the auxiliary equation for the differential equation. $$(r+3)(r-1)(r+5)^3=0$$
Roots are: $r_1=-5$, as a multiplicity of $3$ and $r_2=- 3, r_3=1$ as a multiplicity of $1$.
This implies that there are two independent solutions to the differential equation $y_1(x)=e^{-5x}$ and $y_2=xe^{-5x}$ and $y_3=x^2 e^{-5x}$ and $y_4=e^{-3x}$ and $y_5=e^x$
Therefore, the general equation is equal to $y(x)=C_1e^{-5x}+C_2xe^{-5x}+C_3x^2 e^{-5x}+C_4e^{-3x}+C_5e^x$