Final answer:
To solve initial value problems for the given differential equations, we find the general solution using the characteristic equation and then apply the initial conditions to determine the particular solution that fits those conditions.
Step-by-step explanation:
To solve the given initial value problems for differential equations, we must find the particular solution that satisfies the initial conditions provided. We start by solving the characteristic equation to find the general solution, and then apply the initial conditions to find the constants.
First Initial Value Problem
The characteristic equation for the first problem y'' + 2y' + y = 0; y(0) = 2, y'(0) = -5 is r^2 + 2r + 1 = 0. This is a perfect square (r + 1)^2 = 0, giving a repeated root of r = -1. The general solution is y(t) = C1e-t + C2te-t. Using the initial conditions, we can find C1 and C2.
Second Initial Value Problem
For the second problem y'' + 8y' + 20y = 0; y(0) = 4, y'(0) = -15, we find the characteristic equation r^2 + 8r + 20 = 0, which has the roots r = -4 ± 4i. Therefore, the general solution is y(t) = e-4t(C3cos(4t) + C4sin(4t)). By applying the initial conditions, we solve for C3 and C4.