Fourier transform of derivatives
The single most useful thing the Fourier transform does for differential equations is to make calculus disappear. Differentiation — the operation that makes equations hard — turns, under the transform, into plain multiplication by a number. A differential equation, which relates a function to its derivatives, becomes an algebraic equation, which you solve by dividing.
The rule is short: if f(x) has transform F(k), then the derivative f'(x) has transform i k F(k), and each further derivative brings down another factor of i k, so the n-th derivative transforms to (i k)^n F(k). The mechanism is integration by parts inside the defining integral: differentiating e^{-i k x} produces the factor minus i k, and provided f decays at infinity (so the boundary terms vanish), that factor lands on F(k). Reading it as a picture: a pure wave e^{i k x} differentiated gives i k times itself, so for the wave components the transform diagonalizes differentiation — each frequency is simply scaled by i k.
This is why the Fourier transform solves linear constant-coefficient ordinary and partial differential equations so cleanly. Transform the whole equation, replace every d/dx by i k, and a differential equation in x becomes an algebraic equation in k that you solve for F(k); then invert to recover the solution. For the heat equation, the time derivative stays but the space derivatives become minus k^2, giving F(k, t) = F(k, 0) e^{minus k^2 t} — each Fourier mode just decays at its own rate. The honest condition is that f and enough of its derivatives must vanish at infinity for the boundary terms to drop; on a finite interval with boundary conditions, the Fourier transform's cousins (Fourier series, sine and cosine transforms) take over.
The ODE minus y'' + y = f(x) on the whole line transforms to (k^2 + 1) Y(k) = F(k), so Y(k) = F(k)/(k^2 + 1). The differential equation has become a one-line division; inverting Y(k) (a convolution with the transform of 1/(k^2+1)) gives y(x).
Two derivatives become a factor of k^2, collapsing the ODE into algebra; only the inverse transform requires real work.
The rule (i k)^n F(k) silently assumes the boundary terms from integration by parts vanish — true when f decays at infinity, but false on a bounded interval, where ignoring the boundary contributions gives wrong answers.