FT to solve 2nd order ODE; only one solution

1. Sep 16, 2011

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?

2. Sep 17, 2011

bdforbes

$$f=\frac{1}{2}xsgn(x) + x + 1$$
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.