Separation of variables: when to have exponential solution and when sinusoidal? In separation of variables, one can assume a solution of V(x,y) = X(x)Y(y) and after plugging this into Laplace's equation which is: ${{\partial^2 V} \over {\partial x^2}}$ + ${{\partial^2 V} \over {\partial y^2}}$ = 0 we can get: 
${d^2X \over dx^2}$ = ${k^2X}$ which gives a solution ${X(x) = Ae^{kx} - Be^{-kx}}$? 
and
${d^2Y \over dy^2}$ = ${-k^2Y}$ which gives solution ${Y(y) = C\sin(ky) - D\cos(ky)}$
(where k is some constant) However, I can't understand, why does positive ${k^2}$ give a solution with exponents and ${-k^2}$ has sinusoidal solution? Is it always so?
The book that I am referencing this from is Griffith 3rd edition of "Intro to Electrodynamics". He does mention this there: "If X were sinusoidal, we could never arrange for it to go to zero at infinity, and if Y were exponential we could not make it vanish at both zero and a." 
But I don't quite see this? Also, the image used is: 

Ps: I'm sorry if this has been asked before on this site. If it has, could someone please direct me to the answer, since I could not find it?

For the first question: Yes, those signs will always give those answers. One can check that by direct differentiation. (There's also a physics way to say it: Suppose you had a particle subject to a force which is always directed away from $x=0$. Will the particle oscillate in its position, or will it keep moving away?)
That implies that, as long as $k\neq 0$, one of $X$ or $Y$ will have real exponential solutions and the other sinusoidal. Which goes with which? Recall that sinusoidal functions are periodic, and so can't vanish at infinity. Is this suitable for $Y(y)$?
