Answer
$y= C_{1}e^{\frac{-1}{3}t}+C_{2}e^{\frac{-1}{4} t}+[\;C_{3}e^{- t}cos(2t)+C_{4}e^{-t}sin(2t)\;]$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
$12y^{(4)}+31{y}'''+75{y}''+37{y}'+5y=0 \;\;\;\;\Rightarrow \;\;\;\; 12r^4e^{rt}+31r^3e^{rt}+75r^2e^{rt}+37re^{rt}+5e^{rt}=0\\\\$
$12r^4+31r^3+75r^2+37r+5=(3r+1)(4r+1)(r^2+2r+5)=0 $$\rightarrow\;\;\;\;\; r_{1}= \frac{-1}{3}\;\;\;\;or\;\;\;\;r_{2}=\frac{-1}{4}\;\;\;or\;\;\;r_{3}=-1-2i\;,\;\;r_{4}=-1+2i \;\;\;\;\;\;\\\\$
So the 4 roots are: $\;\;\;r_{1}=\frac{-1}{3} \;\;\;,\;\;r_{2}=\frac{-1}{4} \;\;\;,\;\;\; r_{3},r_{4}=-1\pm 2i$
The general solution for complex roots is:
$y= C_{1}e^{\alpha t}cos(\beta t)+C_{2}e^{\alpha t}sin(\beta t)$
$y= C_{1}e^{\frac{-1}{3}t}+C_{2}e^{\frac{-1}{4} t}+[\;C_{3}e^{- t}cos(2t)+C_{4}e^{-t}sin(2t)\;]$
