I'm having trouble following the solution for this question. Firstly (underlined in orange) why is the general solution of this form. I would have thought it would be the more general $A(w)e^{wy}+B(w)e^{-wy}$ And for the green part I don't really follow it all.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \Phi\pars{x,y}=\int\tilde{\Phi}\pars{\vec{k}}\expo{\ic\vec{k}\cdot\vec{r}}\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}\quad\imp\quad \int\tilde{\Phi}\pars{\vec{k}}\pars{-k_{x}^{2} -k_{y}^{2}} \expo{\ic\vec{k}\cdot\vec{r}}\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}} =0 $$ $\ds{\imp\quad k_{y} = \pm\verts{k_{x}}\ic}$
$$ \Phi\pars{x,y}=\int_{-\infty}^{\infty}{\rm B}\pars{k_{x}} \exp\pars{\ic k_{x}x \color{#c00000}{\Large -} \verts{k_{x}}y}\,{\dd k_{x} \over 2\pi} $$
$$ {\rm p}\pars{x}=\int_{-\infty}^{\infty}{\rm B}\pars{k_{x}} \exp\pars{\ic k_{x}x}\,{\dd k_{x} \over 2\pi}\quad\imp\quad {\rm B}\pars{k_{x}} = \int_{-\infty}^{\infty}{\rm p}\pars{\xi}\expo{-\ic k_{x}\xi}\,\dd\xi $$
\begin{align} \color{#00f}{\large\Phi\pars{x,y}}&=\int_{-\infty}^{\infty}\bracks{% \int_{-\infty}^{\infty}{\rm p}\pars{\xi}\expo{-\ic k_{x}\xi}\,\dd\xi} \expo{\ic k_{x}x - \verts{k_{x}}y}\,{\dd k_{x} \over 2\pi} \\[3mm]&={1 \over 2\pi}\int_{-\infty}^{\infty}{\rm p}\pars{\xi} \int_{-\infty}^{\infty}\expo{\ic k_{x}\pars{x - \xi} - \verts{k_{x}}y}\,\dd k_{x} \,\dd\xi \\[3mm]&={1 \over \pi}\Re\int_{-\infty}^{\infty}{\rm p}\pars{\xi} \int_{0}^{\infty}\!\!\!\!\!\expo{k_{x}\bracks{\ic\pars{x - \xi} - y}}\,\dd k_{x} \,\dd\xi \\[3mm]&={1 \over \pi}\Re\int_{-\infty}^{\infty}{\rm p}\pars{\xi} {-1 \over \ic\pars{x - \xi} - y}\,\dd\xi ={1 \over \pi}\Re\int_{-\infty}^{\infty}{\rm p}\pars{\xi} {\ic\pars{x - \xi} + y \over \pars{x - \xi}^{2} + y^{2}}\,\dd\xi \\[3mm]&=\color{#00f}{\large{y \over \pi}\int_{-\infty}^{\infty} {{\rm p}\pars{\xi} \over \pars{x - \xi}^{2} + y^{2}}\,\dd\xi} \end{align}
