# Inconsistency when solving IVP using Laplace Transform with Dirac Delta

dirac deltainitial-value-problemslaplace transformordinary differential equations

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?

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