Answer
y = $C1cos(\frac{2}{3}x)$ + $C2sin(\frac{2}{3}x)$
Work Step by Step
Question: 9y"+4y=0
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(9r^{2}+4)e^{rx}$=0
The corresponding characteristic equation is:
$9r^{2}+4=0$
Solving the equation with complex numbers gives:
$9r^{2}=-4$
$r^{2}=-\frac{4}{9}$
$r=\frac{+}{-}\sqrt {\frac{4}{9}} i = \frac{+}{-}\frac{2}{3}i$
Adding the constants gives the following general solution:
y = $C1cos(\frac{2}{3}x)$ + $C2sin(\frac{2}{3}x)$
