Final answer:
To solve the initial value problem, we first solve the homogeneous equation, then find the particular solution considering the Dirac delta functions, and finally combine the two solutions to get the complete solution.
Step-by-step explanation:
To solve the initial value problem y'' + 4y' + 13y = δ(t−π/6) + 2δ(t−π/3), y(0) = 1, y'(0) = 2, we will solve the homogeneous equation y'' + 4y' + 13y = 0 first. This will give us the general solution for y. Next, we will find the particular solution by considering the Dirac delta functions. Finally, we will combine the homogeneous and particular solutions to get the complete solution to the initial value problem. Let's solve step by step:
Solving the homogeneous equation:
The characteristic equation for the homogeneous equation is r² + 4r + 13 = 0. Using the quadratic formula, we find that the roots are r = -2 ± 3i. This gives us the solutions y₁(x) = e^(-2x)cos(3x) and y₂(x) = e^(-2x)sin(3x).
Finding the particular solution:
To find the particular solution, we need to consider the Dirac delta functions. Simplifying the equation, we have y'' + 4y' + 13y = δ(t−π/6) + 2δ(t−π/3). Using the properties of the Dirac delta function, we can write the particular solution as y₃(x) = Aδ(x-π/6) + Bδ(x-π/3), where A and B are constants to be determined.
Combining the homogeneous and particular solutions:
We need to find the values of A and B to get the complete solution. To do that, we use the initial conditions y(0) = 1 and y'(0) = 2. Substituting these values into the complete solution, we obtain the following equations: 1 = y₁(0) + y₂(0) + y₃(0) and 2 = y₁'(0) + y₂'(0) + y₃'(0). By solving these equations, we can find the values of A and B, and thus, the complete solution to the initial value problem.
Therefore, the complete solution to the initial value problem y'' + 4y' + 13y = δ(t−π/6) + 2δ(t−π/3), y(0) = 1, y'(0) = 2, is y(x) = y₁(x) + y₂(x) + y₃(x), where y₁(x) = e^(-2x)cos(3x), y₂(x) = e^(-2x)sin(3x), and y₃(x) = Aδ(x-π/6) + Bδ(x-π/3).