Can a real ODE have a complex solution?

Solution 1:

$$ f(x)^2 + 1 = 0 $$ 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 $f(0) = \pm i$.

Solution 2:

$y'' + y = 0$ has solutions $e^{ix}$ and $e^{-ix}$.

Update: $y = ix$ is a solution to $(y')^2 + 1 = 0$ with $y(0)=0$

Solution 3:

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 $y(\sqrt2)=i.$

Solution 4:

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.