FT to solve 2nd order ODE; only one solution

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

However, the general solution is

[tex]f = \frac{1}{2}xsgn(x) + Cx + D[/tex]

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?
 
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My current thinking on this is that we could start with
[tex]f=\frac{1}{2}xsgn(x) + x + 1[/tex]
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.