Word Count : 722

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We are asked to show that the function \[ u = f(x - vt + i\alpha y) + g(x - vt + i\alpha y) \] is a solution of the equation \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}, \] where \(\alpha^2 = 1 - \frac{v^2}{c^2}\). ### Step 1: Find the partial derivatives of \(u\) The function \(u\) is the sum of two terms, \(f(x - vt + i\alpha y)\) and \(g(x - vt + i\alpha y)\), both of which are functions of \(x - vt + i\alpha y\). To compute the required partial derivatives, we'll need to use the chain rule. #### Partial Derivative with Respect to \(x\) For \(f(x - vt + i\alpha y)\), we apply the chain rule: \[ \frac{\partial}{\partial x} f(x - vt + i\alpha y) = f'(x - vt + i\alpha y) \cdot \frac{\partial}{\partial x}(x - vt + i\alpha y) = f'(x - vt + i\alpha y). \] Similarly, \[ \frac{\partial}{\partial x} g(x - vt + i\alpha y) = g'(x - vt + i\alpha y). \] Thus, the total partial _________ ________ ____ _____ ______ _________ _____.
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