Given a solution $u = f(x-c_1t)+g(x-c_2t)$ to $Au_{tt}+2Bu_{tx}+Cu_{xx}=0$ what equation should $c_1$ and $c_2$ satisfy We find a solution to the equation $$Au_{tt}+2Bu_{tx}+Cu_{xx}=0$$ as $$u = f(x-c_1t)+g(x-c_2t)$$ with aribitrary $f,g,$ and real $c_1 < c_2.$ 


*

*What equation should satisfy $c_1$ and $c_2$

*When does this equation have such roots?


So far I have used the solution to rewrite the PDE as 
$$f''(x-c_1t)\left[Ac_1^2 - 2Bc_1 + C\right] + g''(x-c_2t)\left[Ac_2^2 - 2Bc_2 +C\right]=0.$$ I'm stuck on what I ought to do next. I recognize that within each bracket is a quadratic in the respective $c.$ I'm just not sure where to go from here.
 A: Calling 
$$
M = \left(
\begin{array}{cc}
 A & B \\
 B & C \\
\end{array}
\right)
$$
we have
$$
\left(
\begin{array}{c}
 \partial t \\
 \partial x \\
\end{array}
\right)^{\dagger}M\left(
\begin{array}{c}
 \partial t \\
 \partial x \\
\end{array}
\right)u=0
$$
or
$$
\left(
\begin{array}{c}
 \partial t \\
 \partial x \\
\end{array}
\right)^{\dagger}T^{-1}\Lambda T\left(
\begin{array}{c}
 \partial t \\
 \partial x \\
\end{array}
\right)u=0
$$
with
$$
T = \left(
\begin{array}{cc}
 \frac{A-C-\sqrt{4 B^2+(A-C)^2}}{2 B} & 1 \\
 \frac{A-C+\sqrt{4 B^2+(A-C)^2}}{2 B} & 1 \\
\end{array}
\right)\\
\Lambda=\left(
\begin{array}{cc}
 \frac{1}{2} \left(A+C-\sqrt{4 B^2+(A-C)^2}\right) & 0 \\
 0 & \frac{1}{2} \left(A+C+\sqrt{4 B^2+(A-C)^2}\right) \\
\end{array}
\right) = \left(
\begin{array}{cc}
 \lambda_1 & 0 \\
 0 & \lambda_2 \\
\end{array}
\right)
$$
and now calling
$$
\left(
\begin{array}{c}
 \partial \eta \\
 \partial \xi \\
\end{array}
\right) = T\left(
\begin{array}{c}
 \partial t \\
 \partial x \\
\end{array}
\right)
$$
the PDE reduces to
$$
\lambda_1 u_{\eta\eta}+\lambda_2 u_{\xi\xi}=0
$$
and also we can verify by substitution that
$$
u(\eta,\xi) = f(\eta)+g(\xi)
$$
is a solution, or changing variables again
$$
u(t,x) = f\left(\frac{\lambda_1}{B} t + x\right) + g\left(\frac{\lambda_2}{B} t+x\right)
$$
A: You almost got it. If $c_1, c_2$ are zeros of $A x^2 + B x + C$, both terms are zero, whatever $f''(), g''()$ may be. Thus in the proposed solution $f, g$ are arbitrary (but have second derivatives).
