Answer Happy

Given the differential equation y' + 2y' + 3y = 5 sin(4t), y(0) = 2, y'(0) = 1 Apply the Laplace Transform and solve for Y(s) = L[y] Y(s) =
Follow the steps to use the Laplace transform to solve the initial value problem y'' - 3y' + 2y = f(t), y(0) = 0, y'(0) = 0 0 + f(t) = {} where f(t) if t < 2 2t4 if t≥2 a. Write a unit step function for f(t). Use lowercase u for the Heaviside function. f(t) = F₁(s) = F₂(s) = F3(s) = F4(s) = d. Finally, y(t) = b. Find the Laplace transform of the given equation, substitute the initial conditions, and isolate Y(s). Your answer does not have to be in simplest form. Y(s) = c. In simplest form Y(s) can be written as e 3 2s " ( 2² F₁ (8) + F2₂(8) + =—- F3(8) − 2F4(3)) where - - -