## Calculus: Early Transcendentals 8th Edition

$y=e^{3x}(2cos(x)-3sin(x))$
$y''-6y'+10y=0$ 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: $y'(0)=3$ $y'=e^{3x}(-c_{1}sin(x)+c_{2}cos(x))+3e^{3x}(c_{1}cos(3x)+c_{2}sin(3x))$ $y'(0)=3=e^{3(0)}(-c_{1}sin(0)+c_{2}cos(0))+3e^{3(0)}(c_{1}cos(3(0))+c_{2}sin(3(0)))$ $3=c_{2}+3c_{1}$ $3=c_{2}+3(2)$ $c_{2}=-3$ $y=e^{3x}(2cos(x)-3sin(x))$