## Calculus: Early Transcendentals 8th Edition

y = $C1cos(\frac{2}{3}x)$ + $C2sin(\frac{2}{3}x)$
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 = $C1cos(\frac{2}{3}x)$ + $C2sin(\frac{2}{3}x)$