## Calculus 8th Edition

$y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$
The characteristic equation for $y''-6y'+25y=0$ is $t^2-6t+25=0$ with solutions $t=3 \pm 4i$ The general solution is of the form : $c_1e^{3x}cos(4x+c_2e^{3x}sin(4x)$ As per question, $y(0)=2$ and $y'(0)=1$ after plugging in the values, we have $c_1=2$, $c_2=\frac{-5}{4}$ Thus, $y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$