Inconsistency when solving IVP using Laplace Transform with Dirac Delta

Question

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?

Answer 1

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.

1. 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.
2. 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 $$x(t)=(h(t)+1)\,e^{-t}$$