In solving the given initial value problem (IVP) using Laplace transformation, we are provided with the differential equation 0 ≤ t < 3π y" + y = f(t), along with the initial conditions y(0) = 0 and y'(0) = 1. The function f(t) is defined as f(t) = 1.
To solve the given initial value problem (IVP), we can apply the Laplace transformation technique. The Laplace transform allows us to transform a differential equation into an algebraic equation, making it easier to solve. In this case, we have a second-order linear homogeneous differential equation with constant coefficients: y" + y = f(t), where y(t) represents the unknown function and f(t) is the input function.
First, we need to take the Laplace transform of the given differential equation. The Laplace transform of y''(t) is denoted as s^2Y(s) - sy(0) - y'(0), where Y(s) is the Laplace transform of y(t), and y(0) and y'(0) are the initial conditions. Similarly, the Laplace transform of y(t) is Y(s), and the Laplace transform of f(t) is denoted as F(s).
Applying the Laplace transform to the differential equation, we get (s^2Y(s) - sy(0) - y'(0)) + Y(s) = F(s). Substituting the given initial conditions y(0) = 0 and y'(0) = 1, the equation becomes s^2Y(s) - s + Y(s) = F(s).
Now, we can rearrange the equation to solve for Y(s): (s^2 + 1)Y(s) = F(s) + s. Dividing both sides by (s^2 + 1), we find Y(s) = (F(s) + s) / (s^2 + 1).
To find the inverse Laplace transform and obtain the solution y(t), we need to manipulate Y(s) into a form that matches a known transform pair. The inverse Laplace transform of Y(s) will give us the solution y(t) to the IVP.
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