= Solution
\b[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
$$
\boxed{Y(p)=\frac{e^{-p}}{p+1},\qquad\operatorname{Re}p>-1.}
$$
\b[Bromwich inversion.] The <Bromwich inversion formula> is
$$
y(t)=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{p(t-1)}}{p+1}\,dp,\qquad\gamma>-1,
$$
with 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>. Thus
$$
\boxed{y(t)=e^{-(t-1)}\Theta(t-1),}
$$
where $\Theta$ 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.
\b[Fourier transform.] State the convention explicitly:
$$
\widetilde y(k)=\int_{-\infty}^{\infty}y(t)e^{-ikt}\,dt,\qquad y(t)=\frac1{2\pi}\int_{-\infty}^{\infty}\widetilde y(k)e^{ikt}\,dk.
$$
The <Fourier transform> of the distributional <derivative> is $ik\widetilde y(k)$, including the jump at $t=1$. The <Dirac delta> has transform $e^{-ik}$, giving
$$
\boxed{\widetilde y(k)=\frac{e^{-ik}}{1+ik}.}
$$
\b[Inverse Fourier integral.] Substitute into the inverse <Fourier transform>:
$$
y(t)=\frac1{2\pi}\int_{-\infty}^{\infty}\frac{e^{ik(t-1)}}{1+ik}\,dk.
$$
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 $i\operatorname{Res}_{k=i}=i[e^{-(t-1)}/i]=e^{-(t-1)}$. For $t<1$, close in the lower half-plane, where there are no <poles>, giving zero. This reproduces
$$
\boxed{y(t)=\begin{cases}0,&t<1,\\e^{1-t},&t>1,\end{cases}\qquad y(1)=\tfrac12\text{ under symmetric inversion}.}
$$
Integrating the differential equation across the impulse checks the unit jump, $y(1^+)-y(1^-)=1$. Away from the impulse the solution obeys $y'=-y$, and it satisfies the stipulated zero past.
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