Page:Calculus Made Easy.pdf/267

Example 8.

${\frac {d^{2}y}{dt^{2}}}=a^{2}{\frac {d^{2}y}{dx^{2}}}$.

It was seen (p. 177) that this equation was derived from the original

$y=F(x+at)+f(x-at)$,

where $F$ and $f$ were any arbitrary functions of $t$.

Another way of dealing with it is to transform it by a change of variables into

${\frac {d^{2}y}{du\cdot dv}}=0$,

where $u=x+at$, and $v=x-at$, leading to the same general solution. If we consider a case in which $F$ vanishes, then we have simply

$y=f(x-at)$;

and this merely states that, at the time $t=0$, $y$ is a particular function of $x$, and may be looked upon as denoting that the curve of the relation of $y$ to $x$ has a particular shape. Then any change in the value of $t$ is equivalent simply to an alteration in the origin from which $x$ is reckoned. That is to say, it indicates that, the form of the function being conserved, it is propagated along the $x$ direction with a uniform velocity $a$; so that whatever the value of the ordinate $y$ at any particular time $t_{0}$ at any particular point $x_{0}$, the same value of $y$ will appear at the subsequent time $t_{1}$ at a point further along, the abscissa of which is $x_{0}+a(t_{1}-t_{0})$. In this case the simplified