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The given problem is about finding a solution to the one-dimensional wave equation:

\[
\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}
\]

where \(c = 5\). The options provide different functions \(u(x, t)\), and we need to determine which of these functions satisfy the wave equation for the given value of \(c\).

### Step-by-Step Analysis

To determine if a function is a solution to the wave equation, it must satisfy the equation:

\[
\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}
\]

Let's analyze each of the given options:

1. **Option (I): \( u(x, t) = \sin(x + 2t) + \sin(x - 2t) \)**
   - Here, the wave speed would be determined by the coefficients of \(x\) and \(t\) inside the sine functions.
   - For a solution to satisfy the wave equation with \(c = 5\), the coefficient ratio should be such that \( \frac{\text{coefficient of } x}{\text{coefficient of } t} = c = 5 \).
   - For both terms \(\sin(x + 2t)\) and \(\sin(x - 2t)\), the coefficient of \(t\) is 2, and the coefficient of \(x\) is 1. Thus, this does not satisfy \(c = 5\), so **(I) is not a solution**.

2. **Option (II): \( u(x, t) = \sin^2(5x - t) \)**
   - The argument \(5x - t\) suggests a wave speed \(c = \frac{5}{1} = 5\).
   - This matches the given wave speed \(c = 5\). Therefore, **(II) could be a solution**.
   - However, note that this is \(\sin^2(5x - t)\), not a simple sine function. The second derivatives of \(\sin^2\) still satisfy the wave equation due to the chain rule, as it represents a combination of two sinusoidal functions with the correct wave speed.

3. **Option (III): \( u(x, t) = (9x + 5t)^6 \)**
   - The function is a polynomial, not a sinusoidal or cosine function.
   - For this to satisfy the wave equation, it should represent a traveling wave form, which it does not. Therefore, **(III) is not a solution**.

4. **Option (IV): \( u(x, t) = \cos^2(3x - 15t) \)**
   - The argument \(3x - 15t\) suggests a wave speed \(c = \frac{15}{3} = 5\).
   - This matches the given wave speed \(c = 5\). Therefore, **(IV) could be a solution**.
   - The \(\cos^2\) term also satisfies the wave equation similar to the sine squared term due to the chain rule.

### Conclusion

Based on the analysis:

- **(II) and (IV) both satisfy the wave equation for \(c = 5\).**
- Therefore, the correct answer is:

**(II) and (IV) only.**

Would you like to know more about how the wave equation works or how to verify these solutions in detail?

### Related Questions:
1. What is the general solution to the one-dimensional wave equation?
2. How do you determine if a function satisfies a partial differential equation?
3. Why do both sine squared and cosine squared functions satisfy the wave equation?
4. What are some physical interpretations of the wave equation?
5. How would you solve the wave equation using separation of variables?

**Tip:** Understanding the form of solutions to the wave equation is key in physics and engineering, as it represents many physical phenomena like sound and light waves.

Which of the following is a solution of the one-dimensional wave equation ∂²u/∂x² = (1/c²)∂²u/∂t², with c = 5, for the given functions u(x, t)?

This article explores which functions satisfy the one-dimensional wave equation ∂²u/∂x² = (1/c²)∂²u/∂t² for c = 5. Step-by-step analysis verifies the wave equation's solutions using trigonometric and polynomial functions. Recommended for students in advanced high school or early university mathematics courses.

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