Partial

$z = (x - A)^2 + (y - B)^2$

Therefore $\frac{\partial z}{\partial x} = 2(x - A)\;\;\;\;and\;\;\;\;\frac{\partial z}{\partial y} = 2(y - B)$ Note : This equation is of the first order although the equation from which it is derived has two arbitrary constants.

Eliminate A and p from $\displaystyle z = A\,e^{pt}\;sin\;px$

From ths equation it can be seen that:

$\frac{\partial ^2z}{\partial x^2}\;= - p^2\,Ae^{pt}\;sin\;px$ And $\frac{\partial ^2z}{\partial t^2} = p^2\,Ae^{pt}\;sin\;px$

Therefore $\frac{\partial ^2z}{\partial x^2} + \frac{\partial ^2z}{\partial t^2} = 0$

The following linear equation gives the conduction of heat in one direction.

$\frac{\partial ^2z}{\partial x^2} = \frac{1}{a^2}\;\frac{\partial z}{\partial t}$

In the treatment of ordinary linear equations it was found that the use of exponentials was useful. This suggests $\displaystyle z = e^{mx+nt}$ as a trial solution. Substituting in the differential equation: $m^2\;e^{mx+nt} = \frac{1}{a^2}\;ne^{mx+nt}$

Which is true if $\displaystyle n = m^2\,a^2$

Thus $\displaystyle e^{mx+m^2a^2t}$ is a solution. Changing the sign of m gives$\displaystyle e^{-mx+m^2a^2t}$ which is also a solution.

Find a solution to the same differential equation as in example 7 that vanishes when $\displaystyle t = +\,\infty$ . In the previous solutions t occurs in $\displaystyle e^{m^2a^2t}$. This increases with t since $\displaystyle m^2a^2$ is positive if m and a are real. To make it decrease it is necessary to put $\displaystyle m = i\,p$ so that $\displaystyle m^2\;a^2\;= - p^2\;a^2$.

As in the above example this gives $\displaystyle e^{ipx\,-\,p^2a^2t}$ as a solution. Similarly $\displaystyle e^{-\,ipx\,-\,p^2a^2t}$ is also a solution.

Hence as the differential equation is linear the following solution can be written:

$e^{-p^2a^2t}\;(A\,e^{-ipx} + B\,e^{ipx})$

As usual this can be replaced by:

$e^{-p^2a^2t}(E\;cos\;px + F\;sin\;px)$