Is the equation in this context really true?

[yjc]

The short version of this question:
Suppose we have the differential equation $\partial^2_t \phi (x,t) = -k \phi (x,t)$. Is it ever true that
$$i \partial_t \phi (x,t) = \sqrt{k} \phi (x,t) $$
In the one-parameter case with $\phi= \phi (t)$, the general solution is
$$\phi = Ae^{-i\sqrt{k}t} + Be^{i\sqrt{k}t}$$
Then we get
$$i \partial_t \phi (t) = A\sqrt{k}e^{-i\sqrt{k}t} - B\sqrt{k}e^{i\sqrt{k}t} \neq \sqrt{k} \phi (t)$$
, so we end up with a sign change on the second term, so the equation never holds. However, a book I'm using suggests that it is true. Am I making a very simple mistake?

The full context of my question is this. We're given
1. $\partial_t^2 \phi = (\vec{\nabla}^2 - m^2) \phi$
2. $i \langle 0 \vert \partial_t \phi \vert \psi \rangle = ... = \langle 0 \vert \sqrt{m^2 - \vec{\nabla}^2} \phi \vert \psi \rangle$
The omitted steps is when the text expands out the full form of the solution $\phi = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p e^{ipx} + a^{\dagger}_p e^{ipx})$

(In case you're wondering, the $t$-dependence is inside the $x = (t,x_1,x_2,x_3)$.

 

[RUber]

If you are only dealing with the wave in one direction, it seems to work just fine:
$\partial_t Ae^{-ikt} = -ikAe^{-ikt}$.
$i \partial_t \phi = k\phi$.

 

[yjc]

If you are only dealing with the wave in one direction, it seems to work just fine

True, except that the text (QFT, Schwatz) explicitly considers the general case with both terms...

The text also took the trouble to expand out the partial derivative on the solution (the part that I omitted above). I fail to see why that is even necessary.