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The problem presented involves solving the wave equation:

\[
\frac{\partial^2 u}{\partial t^2} = 4 \frac{\partial^2 u}{\partial x^2}, \quad x \in (-\infty, \infty), \quad t > 0
\]
with the initial conditions:
\[
u(x, 0) = 0, \quad u_t(x, 0) = x
\]

This is a classic second-order linear partial differential equation (PDE), often known as the 1-dimensional wave equation. The factor of 4 on the right-hand side implies a wave speed of \( c = 2 \), as the general form of the wave equation is:

\[
\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}
\]

### Solution Approach

The solution to the wave equation can be found using **d'Alembert's formula** for the general solution of the 1D wave equation. For \( c = 2 \), d'Alembert's solution is:

\[
u(x, t) = f(x - 2t) + g(x + 2t)
\]
where \( f \) and \( g \) are functions to be determined from the initial conditions.

#### Step 1: Apply the first initial condition \( u(x, 0) = 0 \)
This gives:
\[
u(x, 0) = f(x) + g(x) = 0 \quad \Rightarrow \quad f(x) = -g(x)
\]

#### Step 2: Apply the second initial condition \( u_t(x, 0) = x \)
The time derivative of the solution is:
\[
u_t(x, t) = -2 f'(x - 2t) + 2 g'(x + 2t)
\]
At \( t = 0 \), this becomes:
\[
u_t(x, 0) = -2 f'(x) + 2 g'(x) = x
\]
Using \( f'(x) = -g'(x) \), we substitute:
\[
-2 f'(x) + 2 (-f'(x)) = x \quad \Rightarrow \quad -4 f'(x) = x
\]
Thus, we solve for \( f'(x) \):
\[
f'(x) = -\frac{x}{4}
\]
Integrating this gives:
\[
f(x) = -\frac{x^2}{8} + C
\]
Since \( f(x) = -g(x) \), we also have \( g(x) = \frac{x^2}{8} - C \).

#### Step 3: Form the general solution
The general solution is:
\[
u(x, t) = f(x - 2t) + g(x + 2t)
\]
Substituting for \( f \) and \( g \):
\[
u(x, t) = \left( -\frac{(x - 2t)^2}{8} + C \right) + \left( \frac{(x + 2t)^2}{8} - C \right)
\]
Simplifying:
\[
u(x, t) = \frac{(x + 2t)^2 - (x - 2t)^2}{8}
\]
This expression simplifies further using the difference of squares:
\[
u(x, t) = \frac{(x^2 + 4xt + 4t^2) - (x^2 - 4xt + 4t^2)}{8} = \frac{8xt}{8} = xt
\]

### Final Solution

Thus, the solution to the wave equation is:
\[
u(x, t) = xt
\]

Would you like any further clarifications or additional details on this solution?

### Related Questions:
1. How is the wave speed \( c \) derived from the wave equation?
2. What is d'Alembert's formula, and how is it used to solve wave equations?
3. How do the initial conditions influence the solution of a PDE?
4. Can this method of solving the wave equation be extended to higher dimensions?
5. How would the solution change if different initial conditions were applied?

### Tip:
For wave equations, always check for symmetry in the initial conditions, as it can simplify finding \( f(x) \) and \( g(x) \) in d'Alembert's solution.

Obtain the solution of the wave equation ∂²u/∂t² = 4 ∂²u/∂x² with initial conditions u(x,0) = 0 and u_t(x,0) = x

Learn how to solve the wave equation ∂²u/∂t² = 4 ∂²u/∂x² with given initial conditions using d'Alembert's solution. This problem focuses on finding the general solution of a second-order PDE, suitable for undergraduate mathematics or physics students.

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