= Solution
{c}
Write $y(x,t)=Y(\xi,\eta)$ with $\xi=x-ct$, $\eta=x+ct$. The <chain rule> gives the <differential operators>
$$
\partial_x=\partial_\xi+\partial_\eta,
\qquad
\partial_t=-c\partial_\xi+c\partial_\eta.
$$
Their squared difference is
$$
y_{tt}-c^2y_{xx}=-4c^2Y_{\xi\eta}.
$$
Consequently the <wave equation> becomes $Y_{\xi\eta}=0$. Integrating first in $\eta$ gives $Y_\xi=F'(\xi)$; integrating in $\xi$ gives the general solution
$$
\boxed{y(x,t)=F(x-ct)+G(x+ct).}
$$
The arbitrary twice differentiable <functions> describe right-moving and left-moving <waves>, respectively. Conversely the <chain rule> shows that each such sum satisfies the <wave equation>.
Back to article page
