A long wire occupying the \(x\) axis is initially at rest. The end \(x=0\) is oscillated up and down so that $$ y(0, t)=2 \sin 3 t, \quad t >0 $$ Find the displacement \(y(x, t)\). The initial and boundary conditions are \(y(0, t)=\) \(2 \sin 3 t, y(x, 0)=0, \partial y /\left.\partial t\right|_{t=0}=0 .\) Take Laplace transforms of these conditions and of the wave equation with respect to \(t\) as in Example 1 . Solve the resulting differential equation to get $$ Y(x, p)=\frac{6 e^{-(p / v) x}}{p^{2}+9} $$Use \(L 3\) and \(L 28\) to find $$y(x, t)=\left\\{\begin{array}{cl} 2 \sin 3\left(t-\frac{x}{v}\right), & x < v t \\ 0, & x > v t \end{array}\right.$$

The Laplace transform is an essential tool in solving differential equations, particularly those involving initial and boundary conditions. Its main advantage is transforming a differential equation, which might be complex in its time domain, into an algebraic equation in a new variable, commonly denoted as \(p\).
This transformation simplifies the handling of initial conditions since they become part of the transformed equation.
For wave equations, such as the one in our exercise, the steps are generally:

• Taking the Laplace transform of the wave equation and the given conditions.
• Solving the resulting algebraic equation in the transform domain.
• Applying the inverse Laplace transform to find the physical solution in the time domain.

By following these steps, we can isolate the function \( Y(x, p) \) and later revert to our original function \( y(x, t) \). The Laplace transform of the sine function, which is a common transformation, is given by: \( L \{ \sin(at) \} = \frac{a}{p^2 + a^2} \). This property was directly used in our exercise.

Initial and boundary conditions define the state of the system at a particular time and location, respectively. They are crucial for finding a unique solution to differential equations.
In our problem, the conditions are given as: \[ y(0, t) = 2 \sin 3 t, \quad y(x, 0) = 0, \quad \frac{\partial y}{\partial t}\Bigg|_{t=0} = 0. \] Here,

• The function \( y(0, t) = 2 \sin 3 t \) describes the displacement at the end of the wire (where \( x \)=0) as time \( t \) progresses.
• The condition \( y(x, 0) = 0 \) states that there is no initial displacement throughout the wire.
• The partial derivative \( \frac{\partial y}{\partial t}\Bigg|_{t=0} = 0 \) specifies that the wire starts from rest, meaning no initial velocity.

These conditions translate to initial values in the Laplace domain, making our transformed equation easier to solve. Proper handling of these conditions ensures an accurate physical representation of the wave propagating through the wire.

The inverse Laplace transform is the process of converting a function from the Laplace domain back to the time domain. Once we solve the algebraic equation in the Laplace domain, we need to apply the inverse transform to recover our solution in original variables.
From our problem solution, we obtained
\( Y(x, p) = \frac{6 e^{-(p/v)x}}{p^2 + 9} \).
To find \( y(x, t) \), we used known inverse Laplace transform properties. Specifically, properties \(L3\) and \(L28\) helped us determine the final form of \( y(x, t) \), which is:
\ y(x, t) = \left\{ \begin{array}{ll} 2 \sin 3\left(t - \frac{x}{v} \right), & x < vt \ 0, & x > vt \end{array} \right. \.
This represents the displacement of the wire at any point \(x\) and time \(t\). Notice how the solution is segmented depending on whether \( x < vt \) or \( x > vt \), which signifies the wave traveling along the wire. Thus, the inverse Laplace transform provides the necessary link back to real-world behavior from the algebraic abstraction.