Final answer:
To solve the Initial Value Problem y''(x) + 2y'(x) + y(x) = 0 with given initial conditions, we find the characteristic equation, solve for roots, and generate two particular solutions y_1(x) = e^{-x} and y_2(x) = xe^{-x}, which are linearly independent.
Step-by-step explanation:
The Initial Value Problem given is y''(x) + 2y'(x) + y(x) = 0, with initial conditions y(0) = -1 and y'(0) = 1. To solve this problem, we first find the characteristic equation associated with the differential equation,
which is r^2 + 2r + 1 = 0. This equation factors into (r + 1)^2 = 0, which has a repeated root, r = -1. Thus, two linearly independent particular solutions to the differential equation are y_1(x) = e^{-x} and y_2(x) = xe^{-x}.
To verify that these solutions satisfy the initial conditions, we substitute x = 0 into both y_1(0) and y_2(0), as well as their first derivatives.