Can I use Dirac $\delta$ as forcing function of a linear differential equation? $$\begin{cases}y'+y=\delta(t) \\ y(0)=0\end{cases}$$
I have used Laplace transform:
$$\mathscr{L} \{ \delta(t) \}=1$$
$$\mathscr{L} \{ y'+y \}=sY(s)-0 +Y(s)=sY(s)+Y(s)$$


$$sY(s)+Y(s)=1$$
$$(s+1) Y(s)=1$$
$$Y(s)=\frac{1}{s+1}$$
Inverse Laplace transform:
$$y(t)=e^{-t}$$
But, $y(0)=1 \ne 0 $

Where is the problem?

Thanks!
 A: Your solution is mostly fine.
The only thing you have to remember is that pointwise value of $y(t)$, especially at $t = 0$, means almost nothing in this business. This is because it is meaningless to interpret $\delta$ in pointwise sense around $t = 0$. Consequently the initial condition is mathematically not well-defined.
Indeed, since the derivative of $y'$ is mostly driven by the impulse $\delta(t)$ near $t = 0$, you should expect that there is a unit jump at $t = 0$. Now by the same computation as LutzL's solution, the general solution is
$$ y(t) = ce^{-t} + e^{-t}u(t) $$
where $u(t)$ is the unit step function. If we interpret the initial condition as $y(0^-) = 0$, then $c = 0$ and we indeed obtain $y(t) = e^{-t} u(t)$. Other interpretation of the condition $y(0) = 0$ will lead to different answers, however, even though they may lead to the value $c\neq 0$ which is physically not plausible.
A: Hint:
$${\cal L}(y'+y)={\cal L}(\delta(t))$$
$$(s+1){\cal L}(y)={\cal L}(\delta(t))$$
$${\cal L}(y)=\dfrac{1}{s+1}{\cal L}(\delta(t))$$
$$y=e^{-t}*\delta(t)$$
A: Multiply with the integrating factor $e^t$ to get
$$
(e^ty(t))'=e^tδ(t)=e^0δ(t)
$$
and integrate to
$$
e^ty(t)=c+u(t)
$$
Now you want, as the use of the Laplace transform indicates, a causal solution, that is, $y(t)=0$ for $t<0$. This requires $c=0$ so that the solution is
$$
y(t)=e^{-t}u(t).
$$
What $y(0)$ is now depends what value the ramp function $u$ takes at the jump.
