Answer
$y= C_{1}e^t+C_{2}e^{(2-\sqrt{5}) t}+C_{3}e^{(2+\sqrt{5}) t}$
Work Step by Step
Let $\;\;\;\;\;\;y=e^{rt}\\\\$
${y}'''+5{y}''+3{y}'+y=0 \;\;\;\;\;\Rightarrow \;\;\;\;\;\; r^3e^{rt}+5r^2e^{rt}+3re^{rt}+e^{rt}=0\\\\$
$r^3+5r^2+3r+1=(r-1)(r^2-4r-1)=0 $$\rightarrow \;\;\;\;\;\; r_{1}= 1\;\;\;\;\;or\;\;\;\;\;r_{2}=2-\sqrt{5}\;,\;r_{3}=2+\sqrt{5} \;\;\;\;\;\;\\\\$
$y= C_{1}e^t+C_{2}e^{(2-\sqrt{5}) t}+C_{3}e^{(2+\sqrt{5}) t}$
