Can a real ODE have a complex solution? By a real ODE I mean an ordinary differential equation with only real coefficients and the resulting function is a function of a real argument. If such a solution exists, can you give an example?
Edit: To add to this, is it still possible if the initial conditions must also be real?
 A: $y'' + y = 0$ has solutions $e^{ix}$ and $e^{-ix}$.
Update: $y = ix$ is a solution to $(y')^2 + 1 = 0$ with $y(0)=0$
A: Consider $y(x)=(1-x^2)^{3/2}$.
Then $$y'(x)=-\frac322x(1-x^2)^{1/2}=-3x\sqrt[3]{y(x)},$$ with the initial condition $y(0)=1$.
But $y(\sqrt2)=i.$
A: There is always a basis of real solutions for a real linear ODE with constant coefficients having degree greater than 1, thus ruling out the weird examples like a polynomial equation having no derivatives at all.  See Theorem 4.1 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/diffeqdim.pdf.
A: Try
$$\left[\frac{dy}{dx}\right]^2 = -(y^2 + 1)$$
This equation can have no real solution at all. Proof by contradiction: assume $y(x)$ is a real valued solution. Then $\left[\frac{dy}{dx}\right]^2$ is real as well, but that implies $\sqrt{-(y(x)^2 + 1)}$ is real, yet $y(x)^2 + 1 > 0$ always no matter what real function $y(x)$ is, thus $-(y(x)^2 + 1) < 0$ always and so this square root can never be real. Contradiction.
EDIT: I just throw it into Wolfram, and it looks like all solutions may be real valued at some isolated points -- but isolated points is not a real-differentiable function of a real variable!
A: $$
f(x)^2 + 1 = 0
$$
That's a differential equation where the "derivative" coefficient is zero; as it happens, the solution is one of the constant functions $f(x) = \pm i$. 
If you want one with a derivative term, consider
$$
f(x)^2 + 1 = xf'(x)
$$
At $x = 0$, any solution must have $f(0) = \pm i$. 
