The given problem is a fourth-order linear differential equation with initial conditions. We will solve it using the Laplace transform method. The equation is:

$x^{(4)}(t) + x(t) = 0$

with the initial conditions: $x(0) = 0, \quad x'(0) = 0, \quad x''(0) = 0, \quad x^{(3)}(0) = 1$

Step 1: Take the Laplace Transform of the equation

We apply the Laplace transform to both sides of the differential equation. The Laplace transform of a derivative is given by the formula:

$\mathcal{L}\{x^{(n)}(t)\} = s^n X(s) - s^{n-1} x(0) - s^{n-2} x'(0) - \cdots - x^{(n-1)}(0)$

Apply the Laplace transform to both terms:

$\mathcal{L}\{x^{(4)}(t)\} + \mathcal{L}\{x(t)\} = \mathcal{L}\{0\}$

Now for each term:

1. $\mathcal{L}\{x^{(4)}(t)\} = s^4 X(s) - s^3 x(0) - s^2 x'(0) - s x''(0) - x^{(3)}(0)$

   Using the initial conditions: $x(0) = 0, x'(0) = 0, x''(0) = 0, x^{(3)}(0) = 1$, we get:

   $\mathcal{L}\{x^{(4)}(t)\} = s^4 X(s) - 1$

2. $\mathcal{L}\{x(t)\} = X(s)$

Thus, the Laplace transform of the equation becomes:

$s^4 X(s) - 1 + X(s) = 0$

Step 2: Solve for $X(s)$

Now, solve the equation for $X(s)$:

$s^4 X(s) + X(s) = 1$

Factor $X(s)$ on the left-hand side:

$X(s) (s^4 + 1) = 1$

Solve for $X(s)$:

$X(s) = \frac{1}{s^4 + 1}$

Step 3: Inverse Laplace Transform

To find $x(t)$, we need to take the inverse Laplace transform of $X(s)$. The function $\frac{1}{s^4 + 1}$ is a standard Laplace transform form. It corresponds to a solution involving sinusoidal functions.

The inverse Laplace transform of $\frac{1}{s^4 + 1}$ is known to be expressed in terms of sine functions involving roots of unity.

$x(t) = \mathcal{L}^{-1} \left\{ \frac{1}{s^4 + 1} \right\}$

The general solution involves complex exponentials or a combination of trigonometric functions. Specifically:

$x(t) = \sum_{k=1}^{4} A_k e^{i \lambda_k t}$

where $\lambda_k$ are the roots of the equation $s^4 + 1 = 0$, which gives $\lambda_k = e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, e^{7i\pi/4}$.

Finally, you can express this as a combination of sines and cosines.

Do you want more details on the inverse Laplace transform or on how to express the solution? Let me know!

Here are some related questions you might want to consider:

1. How can we find the roots of $s^4 + 1 = 0$ explicitly?
2. How do we decompose complex exponentials into real trigonometric functions?
3. How are initial conditions used in solving differential equations with Laplace transforms?
4. What other types of differential equations can be solved with Laplace transforms?
5. What role do Laplace transforms play in control theory and signal processing?

Tip: The Laplace transform is a powerful tool for solving differential equations, especially when initial conditions are present, as it converts the problem into an algebraic equation.