Differential equation in $\mathcal{S}'$, Fourier method I have to solve that equation with Fourier method: $y'-iy=1+\delta'(x)$
Fourier transform is defined like this: 
$F[\varphi](k)=\int\limits_{-\infty}^{\infty}\varphi(x)e^{ikx}dx$
$F^{-1}[\varphi](x)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\varphi(k)e^{-ikx}dk$, where $\varphi \in \mathcal{S}$.
Applying Fourier method:
$F[y]\cdot(k-1) = k - 2\pi i\delta(k)$
The solution is:
$F[y]=A\delta(k-1)+\frac{k}{k-1+i0}+2\pi i\delta(k)$
Now I want to get $y(x)$ applying inverse transform:
$y(x)=\frac{A}{2\pi}e^{-ix}+i+\delta(x)-ie^{-ix}\theta(x)$
So, I know the correct answer
$y(x)=\frac{A}{2\pi}e^{ix}+i+\delta(x)+ie^{ix}\theta(x)$
but I do not see where I've made mistake.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Notation: $\ds{\mrm{f}\pars{x} =
\int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k}\expo{\ic kx}
\,{\dd k \over 2\pi}\iff
\,\hat{\mrm{f}}\pars{k} =
\int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{-\ic kx}\,\dd k}$. 

\begin{align}
&\mrm{y}'\pars{x} - \ic\,\mrm{y}\pars{x} = 1 +\delta\,'\pars{x}
\implies
\ic k\,\hat{\mrm{y}}\pars{k} - \ic\,\hat{\mrm{y}}\pars{k} =
2\pi\,\delta\pars{k} + \ic k
\\[5mm]
\implies &
\hat{\mrm{y}}\pars{k} = {k - 2\pi\,\delta\pars{k}\ic \over k - 1} =
1 + {1 \over k - 1} + 2\pi\,\delta\pars{k}\ic
\end{align}

\begin{align}
\mrm{y}_{\pm}\pars{x} & =
\int_{-\infty}^{\infty}\bracks{1 + {1 \over k - 1 \pm \ic 0^{+}} + 2\pi\,\delta\pars{k}\ic}\expo{\ic kx}\,{\dd k \over 2\pi} =
\delta\pars{x} + \ic + \expo{\ic x}\int_{-\infty}^{\infty}
{\expo{\ic kx} \over k \pm \ic 0^{+}}\,{\dd k \over 2\pi}
\\[5mm] & =
\delta\pars{x} + \ic + \expo{\ic x}\bracks{%
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic kx} \over k}\,{\dd k \over 2\pi} +
\int_{-\infty}^{\infty}\expo{\ic kx}\bracks{\mp\pi\ic\,\delta\pars{k}}
\,{\dd k \over 2\pi}}
\\[5mm] & =
\delta\pars{x} + \ic + \expo{\ic x}\bracks{%
\int_{0}^{\infty}{2\ic\sin\pars{kx} \over k}\,{\dd k \over 2\pi} \mp
{1 \over 2}\,\ic} =
\delta\pars{x} + \ic + \expo{\ic x}\bracks{%
{\ic \over \pi}\,\mrm{sgn}\pars{x}\,{\pi \over 2} \mp {1 \over 2}\,\ic}
\\[5mm] & =
\delta\pars{x} + \ic + \expo{\ic x}
\bracks{{2\Theta\pars{x} - 1 \mp 1}}{\ic \over 2}
\end{align}

$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
\left\{\begin{array}{rcl}
\ds{\quad\mrm{y}_{-}\pars{x}} & \ds{=} &
\ds{\delta\pars{x} + \ic + \Theta\pars{x}\expo{\ic x}\ic}
\\[3mm]
\ds{\quad\mrm{y}_{+}\pars{x}} & \ds{=} &
\ds{\delta\pars{x} + \ic - \Theta\pars{-x}\expo{\ic x}\ic}
\end{array}\right.}}
$$
