Solving a differential equation with a complex number as a coefficient

Question

I am trying to solve the following differential equation; \begin{equation} y'' - iy = 0 \end{equation} By following the usual method of solving, I get my characteristic equation $\lambda^{2} - i = 0$, which then gives me the general solution of \begin{equation} y = cos(\sqrt{i}x) + sin(\sqrt{i}x) \end{equation} I want to know if there is any way I can remove the $i$ term from inside the brackets as to get real solutions. I have tried using an expression for $e^{ix}$ but cannot cancel out the $i$ and the $\sqrt{i}$

I also tried this with a similar ODE, as shown below, \begin{equation} y'' + iy = 0 \end{equation} Where I got a characteristic equation of $\lambda^{2} + i = 0$, giving me a general solution of \begin{equation} y=e^{i\sqrt{i}x} + e^{-i\sqrt{i}x} \end{equation} However, I am still not sure how to cancel this out so that in cos/sin form I have no complex term inside the brackets.

Boundary conditions do not affect the problem at this stage - I'm simply interested in removing the complex term from the brackets.

I hope this is clear enough - any help would be appreciated.

Answer 1

If $y$ is a real solution, taking the imaginary part of $y'' - iy = 0$ (and changing signs) gives $$y = 0 .$$

Answer 2

You can use:

$$\sqrt{i}=(1+i)/\sqrt{2}$$

then you can use

$$\sin ((1\pm i)x/\sqrt{2}) = \sin (x/\sqrt{2}) \ \cos (ix/\sqrt{2}) \pm \cos (x/\sqrt{2}) \ \sin (ix/\sqrt{2})$$

and then you know that

$$ \sin(ix/\sqrt{2})=i\sinh(x/\sqrt{2})$$ $$ \cos(ix/\sqrt{2})=\cosh(x/\sqrt{2})$$

Then you can perform a similar procedure for $\cos ((1\pm i)x/\sqrt{2})$.