Asking the hardcore Laplace wizards here.
I know that the general Complex valued solution to $$y''+k^2y=0\ \ is \ \ \ y(x)=C_1e^{ikx}+C_2e^{-ikx}$$
However, upon trying to reach the same conclusion using the Laplace method for differential equations: $$(s^2+k^2)Y(s)=sy(0)+y'(0) \rightarrow Y(s)=y(0)\frac{s}{s^2+k^2}+y'(0)\frac{1}{k}\frac{k}{s^2+k^2}$$ and therefore: $$y(x)=y(0)cos(kx)+y'(0)\frac{1}{k}sin(kx) = (\frac{y(0)}{2}+\frac{y'(0)}{2ki})e^{ikx}+(\frac{y(0)}{2}-\frac{y'(0)}{2ki})e^{-ikx}=Ae^{ikx}+A^*e^{-ikx} \in \mathbb{R}$$ which is only REAL VALUED and therefore misses the complex solutions.
I'm assuming I'm discarding some assumption made in the process of solving the differential equation using laplace, but I can't figure out where it all goes wrong. Thanks for any help.
This assumes $y(0)$ and $y'(0)$ are real. Regardless of the method, the solution to this equation with real initial conditions will be real for all $x$.
To see this if $$y = C_1 e^{ikx} + C_2 e^{-ikx}$$
then $$\begin{align}y(0)&= C_1 +C_2\\ y'(0)&=ik(C_1-C_2) \end{align}$$
which has the solution
$$y = y(0)\cos(kx) + \frac{y'(0)}{k}\sin(kx)$$
If you let $y(0),y'(0)\in\mathbb{C}$
then $$\frac{y(0)}{2}+\frac{y'(0)}{2ki}\neq \left(\frac{y(0)}{2}-\frac{y'(0)}{2ki}\right)^*.$$
For example, take $$y(0) = i$$ $$y'(0)=0$$
Then $$\frac{y(0)}{2}+\frac{y'(0)}{2ki}= \frac{i}{2}$$
while
$$\left(\frac{y(0)}{2}-\frac{y'(0)}{2ki}\right)^* = -\frac{i}{2}$$