Solve $\begin{equation}u''+u=\delta(x)\end{equation}.$
Using the Fourier integrals we obtain:
$\mathscr{F}\{u''\}= -(\omega)^2\hat{u}$,
$\mathscr{F}\{u\}= \hat{u}$,
$\mathscr{F}\{\delta\}=\frac{1}{\sqrt{2\pi}}$.
We then have $\begin{equation}-(\omega)^2\hat{u}+\hat{u}=\frac{1}{\sqrt{2\pi}}.\end{equation}$
But how do we go from here?
In the absence of boundary (initial) conditions, the solution of the linear differential equation $$u^{\prime \prime}(x)+u(x)=\delta(x)\tag{1} \label{1}$$ is not unique, but given by the general solution $u_h(x)$ of the homogeneous equation $$u_h^{\prime \prime} (x)+u_h(x)=0 \tag{2}$$ plus a special solution $G(x)$ of the inhomogeneous equation $$G^{\prime \prime}(x) +G(x)=\delta(x), \tag{3}$$ where $G(x)$ is a Green function of the differential operator $d^2\!/dx^2+1$.
The analogous situation is encountered after the Fourier transformation, $$\hat{u}(\omega):=\mathcal{F} \{u\}(\omega)=\int\limits_{-\infty}^\infty\! \!\frac{dx}{\sqrt{2\pi}} \, e^{i \omega x} \,u(x) \tag{4} \label{4}$$ mapping \eqref{1} into the equivalent algebraic equation $$(-\omega^2+1)\,\hat{u}(\omega) = \frac{1}{\sqrt{2\pi}}.\tag{5} \label{5}$$ The general solution $\hat{u}_h(\omega)$ of the homogeneous equation $$(-\omega^2+1)\,\hat{u}_h(\omega) =0 \tag{6} $$ is given by $$\hat{u}_h(\omega) = \sqrt{2 \pi} \left[A \,\delta(\omega-1)+B\,\delta(\omega+1)\right], \tag{7} \label{7}$$ with arbitrary constants $A$, $B$. The non-uniqueness of the Green function is now reflected by the fact that the inverse Fourier transform of $$-\frac{1}{\sqrt{2\pi}} \, \frac{1}{\omega^2-1} \tag{8}$$ is undefined because of the presence of the poles at $\omega =\pm 1$. This can easily be cured by defining $\hat{G}(\omega)$ by moving the poles away from the real axis, corresponding to a certain choice of boundary conditions for $$G(x)= \mathcal{F}^{-1} \{ \hat{G} \} (x)= \int\limits_{-\infty}^\infty \!\! \frac{d \omega}{\sqrt{2\pi}} \, e^{-i \omega x} \, \hat{G} (\omega).\tag{9} \label{9}$$ One possibility is given by choosing $$\hat{G}_1(\omega)= -\frac{1}{\sqrt{2\pi}} \, \frac{1}{(\omega-1+i \epsilon)(\omega+1+i \epsilon)}, \qquad \epsilon \downarrow 0. \tag{10} \label{10} $$ $G_1(x)$ is now easily obtained by employing the residue theorem, leading to $$G_1(x)= \Theta(x) \sin x, \tag{11} \label{11}$$ where $\Theta(x)$ denotes the Heaviside step function. Adding \eqref{11} to the inverse Fourier transform of \eqref{7}, gives the general solution of \eqref{1}: $$u(x) = A e^{-ix}+ B e^{ix}+\Theta(x) \sin x. \tag{12} \label{12}$$ The correctness of \eqref{11} can easily be checked by using $\Theta^\prime(x) = \delta(x)$ and $\delta(x) f(x) = \delta(x) f(0)$.