Answer
$y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$
Work Step by Step
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, $y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$
