Solve Second Order ODE involving Dirac Delta using Laplace Transform

Question

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

Answer 1

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.

Answer 2

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\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}

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

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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.