Solution to invariance equation when deriving Lorentz transformation Requiring that the spacetime interval (1 spatial dimension) between the origin and an event $(t,\, x)$ stays constant under a transformation between reference frames: 
$$c^2t^2-x^2= c^2t'^2- x'^2$$
we want to find a solution to that equation, i.e. express $(t,\, x)$ in terms of $(t',\,x')$. In Landau/Lifshitz II it is stated that the general solution is
$$x= x' \mathrm{cosh} (\psi)+ ct' \mathrm{sinh} (\psi),\quad ct= x' \mathrm{sinh}(\psi) + ct' \mathrm{cosh}(\psi)$$
with $\psi$ being the rotating angle in the $tx$ - plane.
It is clear to me that the equation is then satisfied, but how would one find this solution starting from the given equation and why is it the most general one?
 A: The different types of Lorentz transformations can be categorized using group theory, but in this case, I think one can convince themselves with a bit of algebra.
Let's write our transformation in matrix notation. Representing our event as a column vector, $\begin{pmatrix} ct \\ x \end{pmatrix}$, the transformation you wrote above is given as
$\begin{pmatrix}
\cosh (\psi) & \sinh(\psi) \\
\sinh (\psi) & \cosh(\psi)
\end{pmatrix}$
since 
$\begin{pmatrix}
\cosh (\psi) & \sinh(\psi) \\
\sinh (\psi) & \cosh(\psi)
\end{pmatrix}
\begin{pmatrix} ct \\ x \end{pmatrix} = 
\begin{pmatrix}
ct \cosh(\psi) + x \sinh(\psi)  \\ ct \sinh(\psi) + x \cosh(\psi)
\end{pmatrix}\,.$
More generally, transformations can be represented by 2 x 2 matrices,
$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} ct \\ x \end{pmatrix}\,.$
The spacetime interval is given by
$\begin{pmatrix} ct & x \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\begin{pmatrix} ct \\ x \end{pmatrix} = c^2 t^2 - x^2\,.$
Let's act on this value by a general transformation
$\begin{pmatrix} ct & x \end{pmatrix}
\begin{pmatrix} a & c \\ b & d\end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\begin{pmatrix} a & b \\ c & d\end{pmatrix}
\begin{pmatrix} ct \\ x \end{pmatrix}\,.$
Note that the transformation acts twice - once for each copy of the event. Demanding that this value stay the same is identical to demanding 
$\begin{pmatrix} a & c \\ b & d\end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
\begin{pmatrix} a & b \\ c & d\end{pmatrix}
= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\,.$
Doing out the matrix multiplication, we get the constraints $a^2 - c^2 = 1$, $b^2 - d^2 = -1$, and $ac-bd=0$. Assuming we are rotating in the $tx$ plane, $b$ and $c$ cannot equal zero, and we can rule out simple solutions like $a=\pm 1$, $d = \pm 1$. As far as I can tell, hyperbolic functions are the only functions that satisfy those above constraints.
Note that through this formulation, we've essentially reduced our physics question of Lorentz transformations to a mathematical question of matrices - specifically, what sort of matrices satisfy the last matrix equation. Matrices that do this are part of the indefinite orthogonal group $O(p,q)$ In this case, we have one space and one time component, so our group is $O(1,1)$ - I found a more mathematical discussion of how to get all the members of $O(1,1)$ here, in case the above wasn't sufficiently convincing. 
