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Substitution givesThe inverse Fourier transform is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}\tilde{y}(\omega )e^{i\omega t}d\omega$. For $t<0$, close clockwise in the lower half-plane, where there are no poles; hence $y(t)=0$.

For $t>0$, close counterclockwise in the upper half-plane. The three simple-pole residues of the full inversion integrand areMultiplying their sum by $i$ and combining the conjugate exponential contributions yieldsThe large arcs vanish because the rational factor is $O({\omega }^{-3})$ and the exponential is bounded on the chosen half-plane. At zero, the right-hand value and first derivative are both zero, matching the negative-time branch, so no unwanted delta or delta derivative appears in the second-order equation.

For an independent dynamical check, this is the exponential forcing of an underdamped oscillator with damping $\gamma =1$, natural frequency $\sqrt{5}$, damped frequency $2$, exponent $\beta =2$ and force amplitude $-1$. For positive time its substitution gives $y^{\prime \prime }+2y^{\prime }+5y=-e^{-2t}$.