= Solution
The <Fourier transform of a derivative> gives $(-\omega^2-1)\widetilde y=-\widetilde f$. For $\lambda=1$,
$$
\boxed{\widetilde y(\omega)=\frac1{1+\omega^2}\left[\frac{1-i\omega}{(1-i\omega)^2+4}+\frac{1+i\omega}{(1+i\omega)^2+1}\right].}
$$
This selects the solution decaying at both time infinities. Indeed, the inverse transform of $1/(1+\omega^2)$ is the <one-dimensional modified Helmholtz Green function> $K(t)=e^{-|t|}/2$, and the <convolution theorem> gives $y=K*f$. The printed equation has homogeneous solutions $e^t,e^{-t}$, so these two decay conditions matter; no causal initial-value condition was specified.
For $t<0$, evaluate <Fourier inversion> by closing the contour in the lower half-plane, where $e^{i\omega t}$ decays. The contour is clockwise. The enclosed <simple poles> and <residues> of $\widetilde y$ are
$$
\begin{array}{c|c}
\omega&\operatorname{Res}\widetilde y\\\hline
-i&i/5\\
2-i&(-1+i)/16\\
-2-i&(1+i)/16
\end{array}
$$
Thus inversion equals $-i$ times the sum of the residues of $e^{i\omega t}\widetilde y$, giving
$$
\boxed{y(t)=e^t\left[\frac15+\frac18\cos(2t)-\frac18\sin(2t)\right],\qquad t<0.}
$$
As a check, applying $d^2/dt^2-1$ gives $-e^t\cos2t$. The convolution also gives $y(t)=e^{-t}[\tfrac18+\tfrac15\cos t+\tfrac25\sin t]$ for $t\geq0$; the two pieces have matching values $13/40$ and matching first derivatives $3/40$ at zero.
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