Can the Simple Harmonic Oscillator D.E. be Solved Using Fourier Transform? Can I solve the S.H.O differential equation $y'' + \omega^2y = 0$ using the Fourier Transform? 
I tried but couldn't get anywhere with it. I just need to know if it's not possible or whether I'm making the mistake. 
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Lets
$\ds{\,{\rm y}\pars{t}
     =\int_{-\infty}^{\infty}\,\tilde{\rm y}\pars{\nu}\expo{-\ic\nu t}
     \,{\dd\nu \over 2\pi}}$

Then,
  \begin{align}
\pars{-\nu^{2} + \omega^{2}}\,\tilde{\rm y}\pars{\nu}=0\quad\imp\quad
\tilde{\rm y}\pars{\nu}=A\,\delta\pars{\omega + \nu} + B\,\delta\pars{\omega - \nu}
\end{align}

$\ds{A}$ and $\ds{B}$ are constants. Finally,
$$
\,{\rm y}\pars{t}
=\int_{-\infty}^{\infty}\bracks{%
A\,\delta\pars{\omega + \nu} + B\,\delta\pars{\omega - \nu}}\expo{-\ic\nu t}
\,{\dd\nu \over 2\pi}
={1 \over 2\pi}\,\pars{A\expo{\ic\omega t} + A\expo{-\ic\omega t}}
$$

See Dirac Delta Function.

