= Solution
Put $a=1/L\in(0,1)$. The <Fourier transform of a derivative> turns the differential operator into $k^4+5k^2+4=(k^2+1)(k^2+4)$, so
$$
\boxed{\widetilde y(k)=\frac{2a}{(k^2+a^2)(k^2+1)(k^2+4)}.}
$$
Use the <partial fraction decomposition>
$$
\widetilde y=2a\left[\frac1{(1-a^2)(4-a^2)(k^2+a^2)}
-\frac1{3(1-a^2)(k^2+1)}+\frac1{3(4-a^2)(k^2+4)}\right].
$$
The inverse-transform pair $\mathcal F^{-1}[(k^2+b^2)^{-1}]=e^{-b|x|}/(2b)$ is the <one-dimensional modified Helmholtz Green function>. Hence
$$
\boxed{y(x)=\frac{e^{-a|x|}}{(1-a^2)(4-a^2)}-\frac{a e^{-|x|}}{3(1-a^2)}+\frac{a e^{-2|x|}}{6(4-a^2)}.}
$$
In terms of the original length scale,
$$
\boxed{y(x)=\frac{L^4e^{-|x|/L}}{(L^2-1)(4L^2-1)}-\frac{Le^{-|x|}}{3(L^2-1)}+\frac{Le^{-2|x|}}{6(4L^2-1)}.}
$$
This is the <convolution> of $h$ with the <fourth-order Green kernel with two decay scales>, here $K(x)=e^{-|x|}/6-e^{-2|x|}/12$. It decays at both infinities. Although the individual exponential pieces have cusps, their <derivative> jumps through order three cancel, so the result has no unintended <Dirac delta distribution> at zero. The decaying solution is unique: the homogeneous solutions $e^{\pm x},e^{\pm2x}$ have no nonzero globally decaying combination.
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