Answer
$y=e^{3x}(2cos(x)-3sin(x))$
Work Step by Step
$y''-6y'+10y=0$
Use auxiliary equation
$r^{2}-6r+10=0$
$r=\frac{6±\sqrt (36-4(1)(10))}{2(1)}$
$r=\frac{6±\sqrt (36-40)}{2}$
$r=\frac{6±2i}{2}$
$r=3±i$
$α=3$
$β=1$
$y=e^{αx}(c_{1}cos(βx)+c_{2}sin(βx))$
$y=e^{3x}(c_{1}cos(x)+c_{2}sin(x))$
Given: $y(0)=2$
$y(0)=2=e^{3(0)}(c_{1}cos(0)+c_{2}sin(0))$
$c_{1}=2$
Given: $y'(0)=3$
$y'=e^{3x}(-c_{1}sin(x)+c_{2}cos(x))+3e^{3x}(c_{1}cos(3x)+c_{2}sin(3x))$
$y'(0)=3=e^{3(0)}(-c_{1}sin(0)+c_{2}cos(0))+3e^{3(0)}(c_{1}cos(3(0))+c_{2}sin(3(0)))$
$3=c_{2}+3c_{1}$
$3=c_{2}+3(2)$
$c_{2}=-3$
$y=e^{3x}(2cos(x)-3sin(x))$
