Answer
y = $C_{1}e^{2x}cos(3x)$ + $C_{2}e^{2x}sin(3x)$
Work Step by Step
Question: y"-4y'+13y=0
We know:
y = $e^{rx}$
y' = $re^{rx}$
y" = $r^{2}e^{rx}$
This results in $(r^{2}-4r+13)e^{rx}$=0
The corresponding characteristic equation is:
$r^{2}-4r+13=0$
Solving the characteristic equation with the abc-formula and complex numbers gives:
$\frac{-b\frac{+}{-}\sqrt {b^{2}-4*a*c}}{2*a}$ $\frac{4\frac{+}{-}\sqrt {(-4)^{2}-4*1*13}}{2*1}$
$r=2\frac{+}{-}3i$
Adding the constants gives the following general solution:
y = $C_{1}e^{2x}cos(3x)$ + $C_{2}e^{2x}sin(3x)$
