Question

y'' + 4y'+ 13y=0 ; y(0) =2 ; y(0) = -3

Answer 1

This is a 2nd order homogeneous differential equation.
The indicial equation is
m^2 + 4m +13 = 0
Use the quadratic formula.
m = [-4 +/- (16-52)^0.5]/2 = [-4 +/- 6i]/2 = -2 +/- 3i
The general solution is
$y= e^{-2t} [ a cos3t+ b sin3t ]$
$y' = -2e^{-2t}[acos3t + bsin3t] + e^{-2t}[-3asin3t + 3bcos3t]$

Satisfy initial conditions.
y(0)=2 => a = 2
y'(0)=-3 => -2a + 3b = -3

Therefore,  a=2,  b = 1/3

Solution:
$y=e^{-2t}[2cos3t + \frac{1}{3} sin3t ]$