Question on differential equations with $\delta(x)$ In a course of Electrodynamics I came across a function for electric susceptibility $\chi(\tau)$ given by:
$$\frac{d^2\chi}{d\tau^2}+\gamma \frac{d\chi}{d\tau}+\omega_0^2\chi=\omega_p^2\delta(\tau)$$
subject to the boundary conditions $\chi(\tau<0)=0$ and $\chi(\tau\rightarrow 0)\rightarrow0$, whose solution is,
$$\chi(\tau)=\omega_p^2e^{-\gamma\tau/2}\frac{sin(\nu_0)\tau}{\nu_0} \ , \ \ \ \nu^2_0=\omega_0^2-\gamma^2/4$$
I would like to know how to find the solution by myself. Since they give the solution using Fourier transforms, I was wondering how can we find it directly without using Fourier transforms.
However my main question is how do I operate when I have a differential equation with a Delta function using another method, as I haven't studied so far another type of solution. Could you explain or guide me on how to do this?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The differential equation is equivalent to
$$
\left\{\begin{array}{rcl}
\ds{\totald[2]{\chi\pars{\tau}}{\tau} +
\gamma\,\totald{\chi\pars{\tau}}{\tau}+\omega_{0}^{2}\,\chi\pars{\tau}} & \ds{=} & \ds{0\,,
\quad\tau \not= 0}
\\[2mm]
\ds{\lim_{\epsilon \to 0^{+}}\braces{%
\left.\totald{\chi\pars{\tau}}{\tau}\,\right\vert_{\ \tau\ =\ \epsilon} -
\left.\totald{\chi\pars{\tau}}{\tau}\,\right\vert_{\ \tau\ =\ -\epsilon}}} & \ds{=} & \ds{\omega_{p}^{2}}
\end{array}\right.
$$
The homogeneous equation solution is a linear combination of $\ds{\expo{\ic\Omega_{\pm}\tau}}$ where
$$
\Omega_{\pm} \equiv {\gamma\ic \pm \root{-\gamma^{2} + 4\omega_{0}^{2}} \over 2} =
{1 \over 2}\,\gamma\ic \pm \nu_{0}\,,\quad
\nu_{0} \equiv \root{\omega_{0}^{2} - \pars{\gamma \over 2}^{2}}
$$
$\ds{\chi\pars{\tau}}$ is given by
$$
\chi\pars{\tau} =
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} & \ds{\tau} & \ds{<} & \ds{0}
\\[2mm]
\ds{A\expo{\ic\Omega_{+}\tau} + B\expo{\ic\Omega_{-}\tau}}
& \mbox{if} & \ds{\tau} & \ds{>} & \ds{0}
\end{array}\right.
$$

$\ds{\lim_{\tau \to 0^{+}}\chi\pars{\tau} = 0 \implies B = -A}$
and the $\ds{\totald{\chi\pars{\tau}}{\tau}}$ 'jump' at $\ds{\tau = 0}$ leads to

\begin{align}
&A\pars{\ic\Omega_{+} - \ic\Omega_{-}} = \omega_{p}^{2}
\implies
A = {\omega_{p}^{2} \over 2\nu_{0}\ic}
\\ & \mbox{and}\
A\expo{\ic\Omega_{+}\tau} + B\expo{\ic\Omega_{-}\tau} =
{\omega_{p}^{2} \over 2\nu_{0}\ic}\expo{-\gamma\tau/2}\expo{\ic\nu_{0}\tau} -
{\omega_{p}^{2} \over 2\nu_{0}\ic}\expo{-\gamma\tau/2}\expo{-\ic\nu_{0}\tau}
\end{align}

$$
\implies\quad\bbx{\chi\pars{\tau} =
\left\{\begin{array}{lcl}
\ds{0} & \mbox{if} & \ds{\tau} & \ds{<} & \ds{0}
\\[2mm]
\ds{{\omega_{p}^{2} \over \nu_{0}}\,\expo{-\gamma\tau/2}\sin\pars{\nu_{0}\tau}}
& \mbox{if} & \ds{\tau} & \ds{>} & \ds{0}
\end{array}\right.}
$$
