The problem asks us to show that the function $y(x, t) = \frac{1}{2}[f(x - ct) + f(x + ct)]$ satisfies the wave equation $\frac{\partial^2 y}{\partial t^2} = c^2 \frac{\partial^2 y}{\partial x^2}$, where $c$ is a constant and $f$ is any twice differentiable function.

1. Problem Description

The problem asks us to show that the function $y(x, t) = \frac{1}{2}[f(x - ct) + f(x + ct)]$ satisfies the wave equation $\frac{\partial^2 y}{\partial t^2} = c^2 \frac{\partial^2 y}{\partial x^2}$, where $c$ is a constant and $f$ is any twice differentiable function.

2. Solution Steps

First, let's find the second partial derivative of $y$ with respect to $t$, $\frac{\partial^2 y}{\partial t^2}$.

Let $u = x - ct$ and $v = x + ct$. Then $y = \frac{1}{2}[f(u) + f(v)]$.

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

$\frac{\partial^2 y}{\partial t^2} = \frac{\partial}{\partial t} \left(\frac{c}{2} [-f'(x - ct) + f'(x + ct)]\right) = \frac{c}{2} [-f''(x - ct)(-c) + f''(x + ct)(c)] = \frac{c^2}{2} [f''(x - ct) + f''(x + ct)]$

Next, let's find the second partial derivative of $y$ with respect to $x$, $\frac{\partial^2 y}{\partial x^2}$.

$\frac{\partial y}{\partial x} = \frac{1}{2} [f'(u)\frac{\partial u}{\partial x} + f'(v)\frac{\partial v}{\partial x}] = \frac{1}{2} [f'(x - ct)(1) + f'(x + ct)(1)] = \frac{1}{2} [f'(x - ct) + f'(x + ct)]$

$\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{1}{2} [f'(x - ct) + f'(x + ct)]\right) = \frac{1}{2} [f''(x - ct)(1) + f''(x + ct)(1)] = \frac{1}{2} [f''(x - ct) + f''(x + ct)]$

Now, let's check if the wave equation is satisfied:

$\frac{\partial^2 y}{\partial t^2} = c^2 \frac{\partial^2 y}{\partial x^2}$

$\frac{c^2}{2} [f''(x - ct) + f''(x + ct)] = c^2 \left(\frac{1}{2} [f''(x - ct) + f''(x + ct)]\right)$

$\frac{c^2}{2} [f''(x - ct) + f''(x + ct)] = \frac{c^2}{2} [f''(x - ct) + f''(x + ct)]$

The equation holds true. Therefore, $y(x, t) = \frac{1}{2}[f(x - ct) + f(x + ct)]$ satisfies the wave equation.

3. Final Answer

$y(x, t) = \frac{1}{2}[f(x - ct) + f(x + ct)]$ satisfies the wave equation $\frac{\partial^2 y}{\partial t^2} = c^2 \frac{\partial^2 y}{\partial x^2}$.