General solution to the wave equation proving dependence on $x \pm vt$ I am trying to solve for a general solution to the wave function and demonstrate any solution has the form $f(x,t) = f_L (x+vt) + f_R (x-vt)$
I have used separation of variables f(x,t)=X(x)T(t) to decouple the equations: $\dfrac{d^2X}{dx^2} = k^2 x$ and $\dfrac{d^2T}{dt^2} = k^2 v^2 T$
My general solution looks like $X = \sum_k (c_1 x + c_2 + c_3 e^{kx} + c_4e^{-kx} +c_5 e^{ikx} + c_6 e^{-ikx})$, which, throwing out solutions that aren't square integrable, and taking the limit of k being continuous, looks like this: 
$X(x) =\int \phi(k_1) (e^{ik_1 x} + e^{-ik_1 x})dk_1$ and similarly I have 
$ T = \int \phi(k_2)(e^{ik_2 vt} + e^{-ik_2 vt})dk_2$
Now, I am thinking that preserving the positive and negative exponential solutions is not right, and seems meaningless to me since we integrate over all k. But I have kept them in there because it looks at first glance to me like they may help show that $f(x,t) = f_L (x+vt) + f_R (x-vt)$.
Where I am stuck is this last part. Because $k_1$ and $k_2$ are different variables, I cannot show that $f(x,t) = f_L (x+vt) + f_R (x-vt)$ Further, the product $f(x,t)=X(x)T(t)$ will look like $\int \int dk_1 dk_2 \phi(k_1) \phi(k_2)$, which I think may not be correct. 
My question at this point is how can I show that any solution has the form $f(x,t) = f_L (x+vt) + f_R (x-vt)$?
 A: Hint: Use light cone coordinates. What is the full solution to
$$ \frac{\partial^2 f(x^{+},x^{-})}{\partial x^{+}\partial x^{-}}~=~0~?$$
A: Consider showing that $f_L$ and $f_R$ are themselves solutions no matter what function (of one variable) they are.  
Next consider any particular solution (i.e. with particular boundary conditions).  Then try relating your boundary conditions to a particular $f_L$ and $f_R$ jointly.  
If you have a result about uniqueness, then you know exactly how many different boundary conditions you need to get every possible solution, so you know if you have done it for them all.
A: You can show this by noting that $k_1^2=k_2^2$ and $X(x)T(t)= F(k_1x\pm k_2vt)$.
You can see that by:
$$
{ \partial^2 f \over \partial t^2 } = v^2 { \partial^2 f \over \partial  x^2 }
$$
Is more common take $k_1=k$ and $k_2v=-\omega$. Then you may note that the equation imposes the choices of $k_1$ and $k_2$, or $k$ and $\omega$. The imposes is $k_1^2=k_2^2$ or $k^2=\omega^2 v^2$.
Note that you get a mistake in represent the functions $X(x)$ and $T(t)$. Each orthogonal function inside the integration has a free parameter:
$$
X(x) \rightarrow \int  (\phi^+(k_1)e^{ik_1 x} + \phi^-(k_1)e^{-ik_1 x})dk_1
$$
$$
T(t) \rightarrow \int (\phi^+(k_2)e^{ik_2 vt} + \phi^-(k_2)e^{-ik_2 vt})dk_2
$$
So, because $k^2=\omega^2 v^2$, you need to insert delta functions $\delta(k^2-\omega^2 v^2)$ in the combine integration:
$$
X(x)T(t)  =\int\int (\phi^+(k)e^{ik x} + \phi^-(k)e^{-ik x})
 (\phi^+(\omega)e^{i\omega t} + \phi^-(\omega)e^{-i\omega t})\delta(k^2-\omega^2 v^2)dkd\omega
$$
As you can see in other answers that exist other methods that are more natural than yours, by changing the linear operator in terms of light cone variables. But your method is correct anyway.
