Changes

Solutions to this equation are abundant.  For any function <math>\psi(q)</math> of a single variable <math>q</math>, if we turn it into a function of two variables by substituting <math>q = x - vt</math> (or <math>q = x + vt</math>), then, by the [[chain rule]], we have:

Solutions to this equation are abundant.  For any function <math>\psi(q)</math> of a single variable <math>q</math>, if we turn it into a function of two variables by substituting <math>q = x - vt</math> (or <math>q = x + vt</math>), then, by the [[chain rule]], we have:

:<math>\frac{\partial \psi}{\partial t} = \psi' \frac{\partial}{\partial t}(x - ct) = - c \psi'</math>

:<math>\frac{\partial \psi}{\partial t} = \psi' \frac{\partial}{\partial t}(x - ct) = - c\ \psi'</math>

Taking the derivative again, we get:

Taking the derivative again, we get:

:<math>\frac{\partial^2 \psi}{\partial t^2} = c^2 \psi''</math>

:<math>\frac{\partial^2 \psi}{\partial t^2} = c^2\ \psi''</math>

Similarly:

Similarly:

:<math>\frac{\partial \psi}{\partial x} = \psi' \frac{\partial}{\partial x}(x - ct) = \psi'</math>

:<math>\frac{\partial \psi}{\partial x} = \psi' \frac{\partial}{\partial x}(x - ct) = \psi'</math>