Answer
$y=c_{1}e^{\frac{-3}{4}x}cos(\frac{x}{4})+c_{2}e^{\frac{-3}{4}x}sin(\frac{x}{4})$
Work Step by Step
$8y''+12y'+5y=0$
Use auxiliary equation,
$8r^{2}+12r+5=0$
$8r^{2}+12r+5=0$
$r=\frac{-12±\sqrt ((12)^{2}-4(8)(5))}{2(8)}$
$r=\frac{-12±\sqrt (144-160)}{16}$
$r=\frac{-12±\sqrt (-16)}{16}$
$r=-\frac{3}{4}±\frac{i}{4}$
Theorem 4
$y=c_{1}e^{αx}cos(βx)+c_{2}e^{αx}sin(βx)$
$y=c_{1}e^{\frac{-3}{4}x}cos(\frac{x}{4})+c_{2}e^{\frac{-3}{4}x}sin(\frac{x}{4})$
