 Solution

Laplace transform. Use $Y(p)={\int }_0^{\infty }{e}^{-pt}y(t)dt$. The initial value is zero and the Dirac delta contributes $e^{-p}$, so the transformed equation is $pY+Y=e^{-p}$. Therefore

Bromwich inversion. The Bromwich inversion formula iswith the vertical Bromwich contour traversed upwards and its symmetric improper limit understood. For $t>1$, close to the left, where the exponential decays; the enclosed simple pole at $p=-1$ has residue $e^{-(t-1)}$. For $t<1$, close to the right, which contains no pole. Thuswhere $\Theta$ is the Heaviside step function. This agrees with the Laplace transform time-shift rule. At $t=1$, inversion returns the average $1/2$ of the two one-sided values. The value assigned at that single point does not affect the differential equation in the distributional sense.

Fourier transform. State the convention explicitly:The Fourier transform of the distributional derivative is $ik\tilde{y}(k)$, including the jump at $t=1$. The Dirac delta has transform $e^{-ik}$, giving

Inverse Fourier integral. Substitute into the inverse Fourier transform:The only pole is at $k=i$. For $t>1$, close in the upper half-plane. Jordan's lemma removes the large arc and the residue theorem gives $i{\mathrm{Res}}_{k=i}=i[e^{-(t-1)}/i]=e^{-(t-1)}$. For $t<1$, close in the lower half-plane, where there are no poles, giving zero. This reproducesIntegrating the differential equation across the impulse checks the unit jump, $y(1^{+})-y(1^{-})=1$. Away from the impulse the solution obeys $y^{\prime }=-y$, and it satisfies the stipulated zero past.