= Solution
The adjoint also leaves $E$ invariant, with $L^*a=5a-15b$ and $L^*b=3a-9b$. By <adjoint in weighted coordinates>, its <matrix> in the same <basis> is $\begin{pmatrix}5&3\\-15&-9\end{pmatrix}$. Solving the two <eigenvector> equations gives
$$
\boxed{\lambda=0:\ z_0=3e^{-x}-5e^{-2x};\qquad\lambda=-4:\ z_{-4}=e^{-x}-3e^{-2x}.}
$$
The <basis> $a,b$ is not <orthonormal>, so one must not simply transpose the previous numerical <matrix> without accounting for its <Gram matrix>. Directly transposing the integral <kernel> avoids that error.
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