## Calculus: Early Transcendentals 8th Edition

$y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$
$y''-6y'+25y=0$ which has characteristic equation $t^2-6t+25=0$ with solutions $t=3 \pm 4i$ General solution: $c_1e^{3x}cos(4x+c_2e^{3x}sin(4x)$ Plug in the values: $y(0)=2$ and $y'(0)=1$ giving $c_1=2$, $c_2=\frac{-5}{4}$ Hence, $y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$