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 + (1)/(3) sin3t ]`