Inconsistency when solving IVP using Laplace Transform with Dirac Delta solving $\dot{x}(t) + x(t) = \delta (t) $ Using Laplace transform for
$x(0) = 1$, we get:
$sX(s)-1 + X(s) = 1$
$X(s) = \frac{2}{s+1}$
so, $x(t) = 2e^{-t}$
However, evaluating at t=0,
$x(0) = 2 \neq 1$
This disagrees with the initial condition. What went wrong here?
 A: Too long for a comment
There are two important points in this problem.
If you check the solution that you found, you will see that it does not satisfy the equation:
$$ \Big(\frac{d}{dt}+1\Big)2e^{-t}=0\neq \delta(t)$$
The general solution of an inhomogeneous equation (i.e. equation with non-zero RHS) is the general solution of a homogeneous equation, plus a particular solution of an inhomogeneous equation.

*

*A particular solution of the inhomogeneous equation. We don't set any boundary condition (to meet the boundary condition we will use the arbitrary factor at a the general solution of a homogeneous equation).
Applying the Laplace transformation
$$ sX(s)+X(s)=1 \,\Rightarrow \,X(s)=\frac{1}{1+s}\,\Rightarrow \, x_1(t)=h(t)e^{-t}$$
where $h(t)$ is a step-function ($h(t)=0$ for $t<0$, and $h(t)=1$ for $x\geqslant0$).
Given that $h'(t)=\delta(t)$ and $\,\delta(t)f(t)=\delta(t)f(0)$, this is a particular solution of the initial equation.

*The general solution of the homogeneous equation $x'(t)+x(t)=0\,$ is $\,x_2(t)=Ce^{-t}$, where $C$ is an arbitrary constant. So, the general solution of the initial equation is
$$x(t)=h(t)e^{-t}+Ce^{-t}$$
Applying the boundary condition $x(0)=1$ we see that $C=0$. Therefore, the answer, valid for $t\geqslant 0\,$
$$x(t)=h(t)\,e^{-t}$$
If we put the boundary condition, for example, $x(0)=2$, we get the answer
$$x(t)=(h(t)+1)\,e^{-t}$$
A: The reason that you get $x(0)=2$ is that you both set $x(0^-)=1$ and apply a Dirac $\delta$ that adds a unit step making $x(0^+)-x(0^-)=1$. Thus, $x(0^+)=x(0^-)+\left(x(0^+)-x(0^-)\right) = 1 + 1 = 2.$
This will be clearer if you displace the $\delta$ somewhat and take the differential equation as $\dot{x}(t) + x(t) = \delta(t-\epsilon)$ with $x(0^-)=x_0.$
When $\epsilon>0$ this gives $\left(sX(s)-x_0\right)+X(s)=e^{-\epsilon s}$ i.e.
$$
X(s) 
= \frac{x_0+e^{-\epsilon s}}{s+1}
= \frac{x_0}{s+1} + \frac{e^{-\epsilon s}}{s+1}
$$
so
$$
x(t) 
= x_0 e^{-t} H(t) + e^{-(t-\epsilon)} H(t-\epsilon)
.
$$
This function first has a step of size $x_0$ at $t=0$ and later another step of size $1$ (from the $\delta$ term) at $t=\epsilon$. If you take $x_0=1$ and let $\epsilon\to 0$ then you will get a total step of size $2$ at $t=0$ which is what you have seen.
