Fourier Transform to solve partial differential equation In a Cosmology book that I am reading theres an equation 
$$\frac{\partial^2 \delta \epsilon}{\partial t^2}-A\nabla^2 \delta \epsilon-B \delta \epsilon=0~~(1) $$ where $A$ and $B$  do not depend on spatial coordinates and $\delta \epsilon = \delta \epsilon(\vec{x},t)$.
For the next step the author introduces a Fourier Transform in the form of 
$$\int \delta \epsilon(\vec{x},t) = \int \delta \epsilon_{\vec{k}}(t)e^{i \vec{k} 
\cdot \vec{x}} \frac{d^3k}{(2\pi)^{3/2}}$$
Then says "We obtain a set of independent ODE for the time-dependent Fourier coefficients $\delta \epsilon_{\vec{k}}(t)$":
$$\delta \ddot{\epsilon}_{\vec{k}}+w(k)\delta \epsilon_{\vec{k}} = 0~~(2)$$
Where $k$ is the wavevector and $\epsilon$ is the energy density
1-How Fourier Transform can lead such solution ? 
Its also stated that "The advantage of working in Fourier space is that the equation (1) decomposes into a separate equation for each value of k"
So by doing this for every different wavevector we have different solutions for (2) right ?
 A: For ease of notation, I'm going to call $\delta\epsilon=u$. If we take the Fourier transform in space, then $$(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\left(\partial_t^2 u-A\Delta u-Bu\right)e^{-ix\cdot k}\, dx=0.$$ Since we're integrating in $x$, then time derivatives and $A,B$ (space-independent by assumption), can come out of the integral, so that
$$\partial_t^2(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\, u(x,t) e^{-ix\cdot k}\, dx-A(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\, (\Delta u(x,t)) e^{-ix\cdot k}\, dx-B(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\, u(x,t) e^{-ix\cdot k}\, dx=0.$$ Directly from the definition of the Fourier transform, we see that our equation is now $$\partial_t^2 \hat{u}(k,t)-A\widehat{\Delta u}(k,t)-B\hat{u}(k,t)=0,$$ where $\hat{u}(k,t)$ denotes the Fourier transform of $u$ in the $x$ variable. To analyze the second term, we (formally) integrate by parts to obtain 
\begin{align*}
(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\, (\Delta u(x,t)) e^{-ix\cdot k}\, dx&=-(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\, \nabla u(x,t))\cdot \nabla e^{-ix\cdot k}\, dx\\
&=(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\,  u(x,t) (\Delta e^{-ix\cdot k})\, dx\\
&=(-i|k|)^2(2\pi)^{-3/2}\int\limits_{\mathbb{R}^3}\, u(x,t) e^{-ix\cdot k}\, dx\\
&=-|k|^2\hat{u}(k,t).
\end{align*} Putting this back into our earlier equation finally yields
$$\partial_t^2 \hat{u}(k,t)+(A|k|^2-B)\hat{u}(k,t)=0.$$
