= Solution
Write $E_k=e^{i(kx+k^3t)}$, $I=\int_L\varphi(l)\,d\lambda(l)$ and $A_k=\int_L\varphi(l)/(k+l)\,d\lambda(l)$. Introduce the dressing operator
$$
(\mathcal Fh)(k)=h(k)+iE_k\int_L\frac{h(l)}{k+l}\,d\lambda(l).
$$
The given <linear integral equation> is $\mathcal F\varphi=E$, and $q=-I_x$. The appropriate contour and measure must permit differentiation under these integrals and avoid untreated kernel singularities.
For the specified spatial operator, direct differentiation gives
$$
M_kE_k=qE_k,\qquad
M_k(E_kA_k)=E_k(A_{k,xx}+ikA_{k,x}+qA_k).
$$
Inside the integral, replace $\varphi_{xx}(l)+q\varphi(l)$ by $M_l\varphi(l)+il\varphi_x(l)$. It follows that
$$
A_{k,xx}+ikA_{k,x}+qA_k
=\int_L\frac{M_l\varphi(l)}{k+l}\,d\lambda(l)+iI_x.
$$
Apply $M_k$ to $\varphi+iE_kA_k=E_k$. The last term contributes $-E_kI_x=qE_k$, canceling the right side, so
$$
\boxed{\mathcal F(M\varphi)=0.}
$$
This is precisely the homogeneous integral equation. Uniqueness of the original inhomogeneous solution implies a trivial homogeneous <null space>, hence $M\varphi=0$. This calculation is the spatial identity in <Cauchy-kernel dressing for the KdV equation>.
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