y'' + 4y = 8t^2 if 0 < t < 5, and 0 if t > 5; y(1) = 1 + cos(2), y'(1) = 4 - 2sin(2). Use the Laplace transform to find y.
t-shift, s-shift, unit step function.
The Attempt at a Solution
I have been trying to solve it for hours, but keep getting the wrong solution. Each solution takes about 40 minutes, and i'm wondering if there is a faster way than my method, which is:
y'' + 4y = 8t^2 * [1 - u(t-5)] = 8t^2 - 8t^2 * u(t-5)
i set t = T + 1,
ỹ'' + 4ỹ = 8(T + 1)^2 - 8(T + 1)^2 * u(T - 4)
Then I use Laplace on both sides, and get some very messy expressions. I solve for Ỹ = ℒ(ỹ) on the left. I reduce everything on the right with partial fractions., which takes an insane amount of time. Then i find ỹ by taking the inverse Laplace on both sides. Finally, I substitute T with r - 1. Is this the correct way of doing it?