Final answer:
To solve the given initial value problem y'' + 4y' + 29y = 0, we can use the characteristic equation method. The particular solution to the initial value problem is

Step-by-step explanation:
To solve the given initial value problem y'' + 4y' + 29y = 0, we can use the characteristic equation method. First, we assume a solution of the form y = e^(rt). Substituting this into the differential equation, we get the quadratic equation r^2 + 4r + 29 = 0. Solving this quadratic equation, we find that the roots are complex, r = -2 +/- 5i.
Therefore, the general solution to the differential equation is
determined by the initial conditions.
Now, we can use the given initial conditions y(0) = 3 and y'(0) = -5 to find the particular solution. Plugging these values into the general solution, we get the following equations:
3 = C1
-5 = -2C1 + 5C2
Solving these equations simultaneously, we find that C1 = 3 and C2 = -1. Therefore, the particular solution to the initial value problem is
