FT to solve 2nd order ODE; only one solution

[bdforbes]

If I solve a simple 2nd order ODE using a Fourier transform, I only get one solution. E.g.:
\frac{d^2f}{dx^2}=\delta
(2\pi ik)^2\tilde{f}=1
\tilde{f}=\frac{1}{(2\pi ik)^2}
f = \frac{1}{2}xsgn(x)

However, the general solution is

f = \frac{1}{2}xsgn(x) + Cx + D

Why do I only get one of the solutions? Are the solutions with C and D non-zero not also valid distributions whose second derivatives are the delta distribution?

[bdforbes]

My current thinking on this is that we could start with
f=\frac{1}{2}xsgn(x) + x + 1
but as soon as we take the derivatives, we lose the last two terms, and the Fourier transform is then effectively operating on the C=D=0 solution. So it's not really surprising that we only get one solution from the Fourier transform method.