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?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}
$$ \implies\quad\bbx{\chi\pars{\tau} = \left\{\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{\tau} & \ds{<} & \ds{0} \\[2mm] \ds{{\omega_{p}^{2} \over \nu_{0}}\,\expo{-\gamma\tau/2}\sin\pars{\nu_{0}\tau}} & \mbox{if} & \ds{\tau} & \ds{>} & \ds{0} \end{array}\right.} $$