Let’s delve into the problem step by step:

Part (a): Find the Laplace transform of the solution

We begin by taking the Laplace transform of the given differential equation: $x'' + 4\pi^2 x = 2\pi \delta(t-5), \quad x(0) = 0, \quad x'(0) = 0.$

Using Laplace transform properties:

• $\mathcal{L}\{x''(t)\} = s^2 X(s)$,
• $\mathcal{L}\{x(t)\} = X(s)$,
• $\mathcal{L}\{\delta(t - c)\} = e^{-cs}$,
• Initial conditions $x(0) = 0$ and $x'(0) = 0$ simplify terms related to initial conditions.

The equation in the Laplace domain becomes: $s^2 X(s) + 4\pi^2 X(s) = 2\pi e^{-5s}.$

Factor $X(s)$: $X(s) (s^2 + 4\pi^2) = 2\pi e^{-5s}.$

Solve for $X(s)$: $X(s) = \frac{2\pi e^{-5s}}{s^2 + 4\pi^2}.$

Thus, the Laplace transform of the solution is: $\mathcal{L}\{x(t)\} = \frac{2\pi e^{-5s}}{s^2 + 4\pi^2}.$

Part (b): Obtain the solution $x(t)$

Now, we take the inverse Laplace transform of $X(s)$. Using the inverse Laplace transform formula for a shifted term: $\mathcal{L}^{-1}\left\{\frac{1}{s^2 + a^2}\right\} = \frac{\sin(at)}{a},$ and the property $e^{-cs} F(s) \to f(t-c)u(t-c)$, the solution is: $x(t) = \mathcal{L}^{-1}\left\{\frac{2\pi e^{-5s}}{s^2 + (2\pi)^2}\right\}.$

Compute: