= Solution
Write $Y_\pm=\overline y_\pm$, $y_0=y(0)$ and $d_\pm=y'(0\pm)$. In a common convergence strip, two integrations by parts on each half-line give
$$
\int_0^\infty y''e^{ikx}\,dx=-d_++iky_0-k^2Y_+,\qquad
\int_{-\infty}^0y''e^{ikx}\,dx=d_--iky_0-k^2Y_-.
$$
The <boundary> terms at <infinity> vanish by exponential decay; the <ordinary differential equations> also give decay of the required first <derivatives>. Substituting the two equations for $y''$ yields
$$
(a^2-k^2)Y_+=d_+-iky_0,\qquad(b^2-k^2)Y_-=-d_-+iky_0.
$$
<Continuity> makes the $y_0$ terms cancel on addition, while the unit <derivative> jump gives
$$
(a^2-k^2)Y_++(b^2-k^2)Y_-=1.
$$
Multiply by $-1/((k-a)(k+b))$ and factor the quadratics. Since $\operatorname{Im}(a+b)>0$, their sum is nonzero, and partial fractions give
$$
\boxed{\frac{k+a}{k+b}Y_++\frac{k-b}{k-a}Y_-=\frac1{a+b}\left(\frac1{k+b}-\frac1{k-a}\right).}
$$
This is the <Wiener-Hopf equation> for the two <half-line Fourier transforms>. The plus and minus subscripts refer to support in physical space; their <holomorphic> domains in transform space are determined next.
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