How to apply laplace transform to solve X(s)

x"+4x'+4x = f(t)

at time t=0, x=0 and x'=0

Let L{x(t)} = X(s).

(a) Apply laplace transform to the differential equation to solve for X(s) for a general force, f(t).

(b) If the system receives an impulse at t=1 find the response function x(t)

(c) if a force of magnitude 1 applied to the system at t=1 and is left on for t > 1 find the response function x(t)


p/s: sorry, i don't understand this question. can anyone help me please?

(a) [s^2 X(s) - 0 * s - 0] + 4 (s X(s) - 0) + 4 X(s) = F(s)
==> (s^2 + 4s + 4) X(s) = F(s)
==> X(s) = F(s) / (s + 2)^2.

(b) Here, f(t) = δ(x - 1)
==> L{f(t)} = F(s) = e^(-s).

So, X(s) = e^(-s)/(s + 2)^2.

Inverting yields x(t) = e^(-2(t - 1)) (t - 1) u(t - 1).

(c) This time, f(t) = u(t - 1)
==> F(s) = e^(-s) / s.

So, X(s) = e^(-s)/[s(s + 2)^2] = (1/4) e^(-s) [1/(s+2) - 2/(s+2)^2 + 1/s].

Inverting yields x(t) = (1/4) [e^(-2(t - 1)) - 2 (t - 1) e^(-2(t - 1)) + 1] u(t - 1).

I hope this helps!