Final answer:
The initial value problem y'' + 2y' + y = 0; y(0)=2, y'(0)= -5 is solved by finding the characteristic equation, which yields a repeated root r = -1, thus giving the particular solution y(t) = (2 - 3t)e^(-t).
Step-by-step explanation:
To solve the initial value problem y'' + 2y' + y = 0; y(0)=2, y'(0)= -5, we first need to find the general solution to the homogeneous differential equation. For such a second-order linear homogeneous differential equation with constant coefficients, we assume solutions of the form y = e^(rt), where r is a constant that we need to determine. Substituting this form into the differential equation gives a characteristic equation:
r^2 + 2r + 1 = 0.
This is a perfect square trinomial, which factors as (r + 1)^2 = 0. Thus, the repeated root is r = -1. The general solution for the differential equation with a repeated root is:
y(t) = (C1 + C2t)e^(-t)
where C1 and C2 are constants determined by the initial conditions. Using the given initial conditions:
- y(0) = 2 leads to 2 = C1.
- y'(0) = -5 leads to -5 = -C1 + C2, which, given that C1 = 2, results in C2 = -3.
Substituting C1 and C2 back into the general solution, we have the particular solution:
y(t) = (2 - 3t)e^(-t)