Question

Show that the function satisfies Laplace's equation \(\partial^{2} z / \partial x^{2}+\partial^{2} z / \partial y^{2}=0\). \(z=\frac{1}{2}\left(e^{y}-e^{-y}\right) \sin x\)

Short answer

The function \(z=\frac{1}{2}\left(e^{y}-e^{-y}\right) \sin x\) does satisfy Laplace's Equation

Step-by-step solution

Calculate the second partial derivative with respect to x

The given function is \(z=\frac{1}{2}\left(e^{y}-e^{-y}\right) \sin x\). Differentiating twice with respect to x, first derivative is \(z_x= \frac{1}{2}\left(e^{y}-e^{-y}\right) \cos x\), and the second derivative \(z_{xx}= -\frac{1}{2}\left(e^{y}-e^{-y}\right) \sin x\)

Calculate the second partial derivative with respect to y

Next, we differentiate twice with respect to y. The first derivative is \(z_y= \frac{1}{2}(e^{y}+e^{-y}) \sin x\), and the second derivative is \(z_{yy}= \frac{1}{2}(e^{y}-e^{-y}) \sin x\)

Check if the sum of second partial derivatives is zero

\(\frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2} z}{\partial y^{2}} = z_{xx} + z_{yy} = -\frac{1}{2}\left(e^{y}-e^{-y}\right) \sin x + \frac{1}{2}(e^{y}-e^{-y}) \sin x = 0\)