General form of the solution of a damped wave equation In my physics textbooks, I often see a general solution written for the wave equations. The equation $$\frac{\partial^2\psi(x,t)}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\psi(x,t)}{\partial t^2}\label{1}\tag{1}$$ has the general solution of the form $$\psi(x,t)=f(x+ct)+g(x-ct).\label{1a}\tag{1a}$$ where $f,g$ are arbitrary functions. 
Is it possible to write down a general form of the solution for the equation \eqref{2} $$\frac{\partial^2\psi(x,t)}{\partial x^2}=a\frac{\partial\psi(x,t)}{\partial t}+\frac{1}{c^2}\frac{\partial^2\psi(x,t)}{\partial t^2}\label{2}\tag{2}$$ in a similar form as \eqref{1a}?
 A: Let $\psi(x,t)=e^{-\frac{ac^2t}{2}}u(x,t)$ ,
Then $\dfrac{\partial\psi(x,t)}{\partial t}=e^{-\frac{ac^2t}{2}}\dfrac{\partial u(x,t)}{\partial t}-\dfrac{ac^2e^{-\frac{ac^2t}{2}}u(x,t)}{2}$
$\dfrac{\partial^2\psi(x,t)}{\partial t^2}=e^{-\frac{ac^2t}{2}}\dfrac{\partial^2u(x,t)}{\partial t^2}-\dfrac{ac^2e^{-\frac{ac^2t}{2}}}{2}\dfrac{\partial u(x,t)}{\partial t}-\dfrac{ac^2e^{-\frac{ac^2t}{2}}}{2}\dfrac{\partial u(x,t)}{\partial t}+\dfrac{a^2c^4e^{-\frac{ac^2t}{2}}u(x,t)}{4}=e^{-\frac{ac^2t}{2}}\dfrac{\partial^2u(x,t)}{\partial t^2}-ac^2e^{-\frac{ac^2t}{2}}\dfrac{\partial u(x,t)}{\partial t}+\dfrac{a^2c^4e^{-\frac{ac^2t}{2}}u(x,t)}{4}$
$\dfrac{\partial\psi(x,t)}{\partial x}=e^{-\frac{ac^2t}{2}}\dfrac{\partial u(x,t)}{\partial x}$
$\dfrac{\partial^2\psi(x,t)}{\partial x^2}=e^{-\frac{ac^2t}{2}}\dfrac{\partial^2u(x,t)}{\partial x^2}$
$\therefore e^{-\frac{ac^2t}{2}}\dfrac{\partial^2u(x,t)}{\partial x^2}=ae^{-\frac{ac^2t}{2}}\dfrac{\partial u(x,t)}{\partial t}-\dfrac{a^2c^2e^{-\frac{ac^2t}{2}}u(x,t)}{2}+\dfrac{e^{-\frac{ac^2t}{2}}}{c^2}\dfrac{\partial^2u(x,t)}{\partial t^2}-ae^{-\frac{ac^2t}{2}}\dfrac{\partial u(x,t)}{\partial t}+\dfrac{a^2c^2e^{-\frac{ac^2t}{2}}u(x,t)}{4}$
$\dfrac{1}{c^2}\dfrac{\partial^2u(x,t)}{\partial t^2}=\dfrac{\partial^2u(x,t)}{\partial x^2}+\dfrac{a^2c^2u(x,t)}{4}$
$\dfrac{\partial^2u(x,t)}{\partial t^2}=c^2\dfrac{\partial^2u(x,t)}{\partial x^2}+\dfrac{a^2c^4u(x,t)}{4}$
Let $p=\dfrac{acx}{2}$ ,
Then $\dfrac{\partial^2u(p,t)}{\partial t^2}=\dfrac{a^2c^4}{4}\dfrac{\partial^2u(p,t)}{\partial p^2}+\dfrac{a^2c^4u(p,t)}{4}$
Let $q=\dfrac{ac^2t}{2}$ ,
Then $\dfrac{\partial^2u(p,q)}{\partial q^2}=u(p,q)+\dfrac{\partial^2u(p,q)}{\partial p^2}$
Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics
Consider $u(p,a)=f(p)$ and $u_q(p,a)=g(p)$ ,
Let $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^n}{n!}\dfrac{\partial^nu(p,a)}{\partial q^n}$ ,
Then $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(p,a)}{\partial q^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(p,a)}{\partial q^{2n+1}}$
$\dfrac{\partial^4u(p,q)}{\partial p^4}=\dfrac{\partial^2u(p,q)}{\partial p^2}+\dfrac{\partial^4u(p,q)}{\partial q^2\partial p^2}=u(p,q)+\dfrac{\partial^2u(p,q)}{\partial q^2}+\dfrac{\partial^2u(p,q)}{\partial q^2}+\dfrac{\partial^4u(p,q)}{\partial q^4}=u(p,q)+2\dfrac{\partial^2u(p,q)}{\partial q^2}+\dfrac{\partial^4u(p,q)}{\partial q^4}$
Similarly, $\dfrac{\partial^{2n}u(p,q)}{\partial p^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}u(p,q)}{\partial q^{2k}}$
$\dfrac{\partial^3u(p,q)}{\partial p^3}=\dfrac{\partial u(p,q)}{\partial p}+\dfrac{\partial^3u(p,q)}{\partial q^2\partial p}$
$\dfrac{\partial^5u(p,q)}{\partial p^5}=\dfrac{\partial^3u(p,q)}{\partial p^3}+\dfrac{\partial^5u(p,q)}{\partial q^2\partial p^3}=\dfrac{\partial u(p,q)}{\partial p}+\dfrac{\partial^3u(p,q)}{\partial q^2\partial p}+\dfrac{\partial^3u(p,q)}{\partial q^2\partial p}+\dfrac{\partial^5u(p,q)}{\partial q^4\partial p}=\dfrac{\partial u(p,q)}{\partial p}+2\dfrac{\partial^3u(p,q)}{\partial q^2\partial p}+\dfrac{\partial^5u(p,q)}{\partial q^4\partial p}$
Similarly, $\dfrac{\partial^{2n+1}u(p,q)}{\partial p^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}u(p,q)}{\partial q^{2k}\partial p}$
$\therefore u(p,q)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(p)(q-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(p)(q-a)^{2n+1}}{(2n+1)!}$
