Solving an electromagnetism problem, I came across this equation:
$$ \nabla^2 B = c^2 B $$
where $c \in \mathbb{R}$ and $B$ is a 3-dimensional vector field. Apparently its solutions are supposed to be of the form $$ C_1 e^{ax} + C_2 e^{-ax}$$ I haven't really taken a course in PDEs yet so I am puzzled by this. Do you have any idea how I can show this is true?
I guess to get a solution like the one you have given you can factor the PDE like a second order ODE: $$ (\nabla - c)(\nabla + c)\cdot B = 0 $$ which gives you the two PDEs: $$ \nabla\cdot B = \sum_k \frac{\partial B}{\partial x_k} = c \quad \text{and} \quad \nabla\cdot B = \sum_k \frac{\partial B}{\partial x_k} = -c $$ These two equations admit a lot of solutions, if you say that $B$ is a function of a single variable, you get something like what you have in your question since the gradient collapses to a univariate total derivative. But without more conditions there isn't some straightforward way to solve this system that I know of.