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The question provided in the image is asking to prove a partial differential equation:

**Given:**

\[
u = f(x - ct) + g(x + ct)
\]

Prove that:

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

### Solution Outline:

1. **First Partial Derivative with respect to \(x\):**
   - \( u(x,t) = f(x - ct) + g(x + ct) \)
   - Use the chain rule to differentiate both \( f(x - ct) \) and \( g(x + ct) \) with respect to \(x\).

2. **Second Partial Derivative with respect to \(x\):**
   - Differentiate the result from step 1 again with respect to \(x\).

3. **First Partial Derivative with respect to \(t\):**
   - Apply the chain rule to differentiate both \( f(x - ct) \) and \( g(x + ct) \) with respect to \(t\).

4. **Second Partial Derivative with respect to \(t\):**
   - Differentiate the result from step 3 again with respect to \(t\).

5. **Equating the second partial derivatives:**
   - Show that \( \frac{\partial^2 u}{\partial x^2} \) equals \( \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} \).

Let’s go through the process step by step:

### 1. First Partial Derivative with respect to \(x\):

\[
\frac{\partial u}{\partial x} = \frac{d}{dx}[f(x - ct)] + \frac{d}{dx}[g(x + ct)]
\]

Since \( \frac{d}{dx}(x - ct) = 1 \) and \( \frac{d}{dx}(x + ct) = 1 \), we get:

\[
\frac{\partial u}{\partial x} = f'(x - ct) + g'(x + ct)
\]

### 2. Second Partial Derivative with respect to \(x\):

\[
\frac{\partial^2 u}{\partial x^2} = \frac{d}{dx}[f'(x - ct)] + \frac{d}{dx}[g'(x + ct)]
\]

Again, differentiating with respect to \(x\):

\[
\frac{\partial^2 u}{\partial x^2} = f''(x - ct) + g''(x + ct)
\]

### 3. First Partial Derivative with respect to \(t\):

\[
\frac{\partial u}{\partial t} = \frac{d}{dt}[f(x - ct)] + \frac{d}{dt}[g(x + ct)]
\]

Now, \( \frac{d}{dt}(x - ct) = -c \) and \( \frac{d}{dt}(x + ct) = c \), so:

\[
\frac{\partial u}{\partial t} = -c f'(x - ct) + c g'(x + ct)
\]

### 4. Second Partial Derivative with respect to \(t\):

\[
\frac{\partial^2 u}{\partial t^2} = \frac{d}{dt}[-c f'(x - ct)] + \frac{d}{dt}[c g'(x + ct)]
\]

Differentiating again:

\[
\frac{\partial^2 u}{\partial t^2} = c^2 f''(x - ct) + c^2 g''(x + ct)
\]

### 5. Equating the Second Partial Derivatives:

From step 2, we have:

\[
\frac{\partial^2 u}{\partial x^2} = f''(x - ct) + g''(x + ct)
\]

From step 4, we have:

\[
\frac{\partial^2 u}{\partial t^2} = c^2 f''(x - ct) + c^2 g''(x + ct)
\]

Now divide both sides of the second equation by \(c^2\):

\[
\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = f''(x - ct) + g''(x + ct)
\]

Thus:

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

This proves the required equation.

---

Would you like a deeper explanation of any of the steps, or have further questions? Here are some related topics:

1. What is the wave equation and how is it related to this problem?
2. How does the chain rule apply in multivariable calculus?
3. What is the physical interpretation of this equation in wave mechanics?
4. Can this method be extended to other types of wave equations?
5. How would this proof change if different functions for \(f\) and \(g\) were used?

**Tip:** The key to solving these kinds of problems is carefully applying the chain rule when differentiating functions of multiple variables.

Jika u = f(x - ct) + g(x + ct), tunjukkan bahwa: ∂²u / ∂x² = (1 / c²) ∂²u / ∂t²

This problem involves proving a partial differential equation related to wave motion, where u(x, t) = f(x - ct) + g(x + ct). The solution uses the chain rule to compute second derivatives with respect to x and t, demonstrating that ∂²u / ∂x² = (1 / c²) ∂²u / ∂t².

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