= Solution
The nonzero <kernel> of $L$ rules out a unique solution for every value of $c$. By the <Fredholm alternative>, solvability requires <orthogonality> of the forcing to $\ker L^*=\operatorname{span}\{3a-5b\}$. Its pairing with $6a+cb$ is
$$
\langle3a-5b,6a+cb\rangle=9+c-10-\frac{5c}4=-1-\frac c4.
$$
Thus only $c=-4$ is compatible. In that case $6a-4b=2y_{-4}$ and a particular solution is $-y_{-4}/2$. All solutions are
$$
\boxed{y(x)=-\frac32e^{-x}+e^{-2x}+C(5e^{-x}-6e^{-2x}),\qquad c=-4.}
$$
Therefore \b[infinitely many solutions exist for $c=-4$, no solution exists for $c\ne-4$, and no parameter value gives a unique solution].
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