Question

y'' +4π²ʸ = 2πδ(t−5), y(0)= 0, y'(0) =0. Find the Laplace transform of the solution: Y(s) = L{y(t)}= Obtain the solution y(t) using the Laplace transform: y(t) = ? Please proceed with solving both parts of the question.

Answer 1

To find the Laplace transform of the given differential equation y'' + 4π²ʸ = 2πδ(t−5), we apply the Laplace transform to both sides of the equation. Taking the inverse Laplace transform of the resulting equation gives us the solution y(t). The solution is y(t) = (πe^(2πt) - πe^(-2πt))u(t-5), where u(t-5) is the unit step function.

Step-by-step explanation:

To solve the given differential equation using the Laplace transform, we first need to take the Laplace transform of both sides of the equation. Applying the Laplace transform to y'' + 4π²ʸ, we get s²Y(s) - sy(0) - y'(0) + 4π²Y(s) = 2πe^(-5s). Substituting the initial conditions y(0) = 0 and y'(0) = 0, we get s²Y(s) + 4π²Y(s) = 2πe^(-5s). Rearranging the equation, we have Y(s) = 2πe^(-5s) / (s² + 4π²).

To find the solution y(t), we need to take the inverse Laplace transform of Y(s). Using partial fraction decomposition or the method of residues, you can find that the inverse Laplace transform of Y(s) is y(t) = (πe^(2πt) - πe^(-2πt))u(t-5), where u(t-5) is the unit step function.