Solve Second Order ODE involving Dirac Delta using Laplace Transform My problem is to solve
$$
\frac{\mathrm{d}^{2}x}{\mathrm{d}{t}^2} + 5\,\frac{dx}{dt} + 6x =
\delta\left(t\right)\quad\mbox{with}\quad x\left(0\right) = 0\quad
\mbox{and}\quad
\left.\frac{\mathrm{d}x}{\mathrm{d}{t}}\right\vert_{\ t\ =\ 0} = 0
$$
Taking the Laplace transform of both sides, I get
$(s^2+5s+6)L[x(t)]=L[\delta(t)]=1$
$L[x(t)]=\frac{1}{(s+2)(s+3)}=L[e^{-2t}-e^{-3t}]$
$x(t)=e^{-2t}-e^{-3t}$.
But this does not satisfy the equation or initial conditions
Could anyone help please? Thank you
 A: Your solution is $x(t) = (e^{-2t}-e^{-3t})1_{t > 0}$ that is correct for the initial condition $x(t) =  x'(t) = 0 $ for $t < 0$.
Note that the product rule (in the sense of distributions) applies, because it is the product of a distribution ($1_{t > 0}$) and a $C^\infty$ function, i.e. infinitely differentiable, ($e^{-at}$) :  $$\frac{d}{dt} [e^{-at}1_{t > 0}] = 1_{t > 0}\frac{d}{dt} e^{-at}+e^{-at}\frac{d}{dt} 1_{t > 0} = - e^{-at}1_{t > 0}+ e^{-at}\delta(t) = - e^{-at}1_{t > 0}+ \delta(t)$$ hence $$x'(t) = -(2e^{-2t}-3e^{-3t})1_{t > 0}, \qquad x''(t) = (4e^{-2t}-9e^{-3t})1_{t > 0} +\delta(t)$$
and you can check that $x''(t)+5x'(t)+6x(t)= \delta(t)$.
Now since it is a linear differential equation, all the other solutions are of the form $\tilde{x}(t) = x(t)+y(t)$ where $y(t)$ is a solution of the homogeneous equation $y''(t)+5y'(t)+6y(t)= 0$. The general solution of this being $y(t) = Ae^{-2t}+Be^{-3t}$ you get that the general solution of $\tilde{x}''(t)+5\tilde{x}'(t)+6\tilde{x}(t)= \delta(t)$ is 
$$\tilde{x}(t) = (e^{-2t}-e^{-3t})1_{t > 0}+Ae^{-2t}+Be^{-3t}$$
and none of them is $C^1$ at $t=0$, hence as stated your exercice is incorrect.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{align}
&\totald[2]{\,\mrm{x}\pars{t}}{t} + 5\,\totald{\,\mrm{x}\pars{t}}{t} + 6\,\mrm{x}\pars{t} =
\delta\left(t\right)\label{1}\tag{1}
\end{align}

$$
\begin{array}{|l|}\hline \mbox{}\\
\mbox{The OP asks for the boundary conditions}\ \ds{\,\mrm{x}\pars{0} = 0}\
\mbox{and}\ \require{cancel}
\cancel{\ds{\left.\totald{\mrm{x}\pars{t}}{t}\right\vert_{\ t\ =\ 0} = 0}}.
\mbox{However,}
\\
\\\mbox{the boundary condition which involves the derivative at}\ 
\ds{t = 0}\ \mbox{doesn't make any sense}
\\
\mbox{because the}\ Dirac\ Delta\ \mbox{enforces the condition}
\\[2mm]
\ds{\left.\totald{\mrm{x}\pars{t}}{t}\right\vert_{\ t\ =\ 0^{+}} -
\left.\totald{\mrm{x}\pars{t}}{t}\right\vert_{\ t\ =\ 0^{-}} \equiv
\left.\totald{\mrm{x}\pars{t}}{t}
\right\vert_{\ t\ =\ 0^{-}}^{\ t\ =\ 0^{+}} = 1}
\quad\mbox{such that}\
\ds{\totald{\mrm{x}\pars{t}}{t}}\ \mbox{is not defined at}\ \ds{t = 0}.
\\ \mbox{}\\ \hline
\end{array}
$$

Solutions of the homogeneous equation $\ds{\totald[2]{\,\mrm{x}\pars{t}}{t} + 5\,\totald{\,\mrm{x}\pars{t}}{t} + 6\,\mrm{x}\pars{t} = 0}$ are $\ds{\expo{-2t}}$ y $\ds{\expo{-3t}}$

The general solution which is continuos at $\ds{t = 0}$ and satisfies $\ds{\,\mrm{x}\pars{0} = 0}$ is given by:
\begin{align}
\mrm{x}\pars{t} & =
\left\{\begin{array}{lcl}
\ds{A\pars{\expo{-2t} - \expo{-3t}}} & \mbox{if} & \ds{t < 0}
\\[2mm]
\ds{B\pars{\expo{-2t} - \expo{-3t}}} & \mbox{if} & \ds{t > 0}
\end{array}\right.
\end{align}
where $\ds{A}$ and $\ds{B}$ are $\ds{t}$-independent constants.

The condition
$\ds{\left.\totald{\,\mrm{x}\pars{t}}{t}
\right\vert_{\ t\ =\ 0^{-}}^{\ t\ =\ 0^{+}} = 1\implies
B\bracks{-2 - \pars{-3}} - A\bracks{-2 - \pars{-3}} = 1\implies\ B = A + 1}$:
$$\bbox[15px,#ffe,border:0.1em groove navy]{%
\mrm{x}\pars{t} =
\left\{\begin{array}{lcl}
\ds{A\pars{\expo{-2t} - \expo{-3t}}} & \mbox{if} & \ds{t < 0}
\\[2mm]
\ds{\pars{A + 1}\pars{\expo{-2t} - \expo{-3t}}} & \mbox{if} & \ds{t > 0}
\end{array}\right\}
=
\bracks{A + \Theta\pars{t}}\pars{\expo{-2t} - \expo{-3t}}} 
$$

$\ds{\Theta}$ is the Heaviside Step Function.

