If $u : \Bbb R \to \Bbb R$ satisfies $u' + 2\pi x u = 0$, why does $\hat{u}$ (the Fourier transform) also satisfy this?

Question

I'm trying to understand why if a function $u : \Bbb R \to \Bbb R$ satisfies the differential equation $u' + 2\pi x u = 0$, then so does the Fourier transform.

The properties I have that I can use are: $\widehat{\frac{du}{dx}}(\xi) = 2 \pi i \xi \hat{u}(\xi)$, and also $-\widehat{2 \pi i x u}(\xi) = \frac{d\hat{u}}{d\xi}(\xi)$.

I think this is supposed to be really simple, but I've tried substituting the second property with no luck.

Answer 1

We have: $$ u' = -2\pi x u, $$ so transforming both sides we have: $$ 2\pi i\xi\, \widehat{u}(\xi) = -i \widehat{u}'(\xi).$$