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## Homework Statement

Find the solution to:

$$\frac{d^2}{dt^2} x + \omega^2 x = \delta (t)$$

Given the initial condition that ##x=0## for ##t<0##. First find the general solution to ##t>0## and ##t<0##.

## Homework Equations

## The Attempt at a Solution

This looks like a non-homogeneous second order DE since it has the form ##ax''+bx'+cx=G(t)##. I think the general solution to this kind of DE is of the form:

$$x(t) = x_p(t)+x_c (t)$$

where x

_{p}is a particular solution of the equation, and x

_{c}is the general solution of the complementary equation ##ax''+bx'+cx=0##.

So, to find x

_{c}we can find the roots of the auxiliary equation of ##x''+\omega^2 x =0##:

$$r^2+\omega^2 = 0 \implies r=\sqrt{-\omega^2} = j \omega$$

Therefore ##x_c = c_1 e^{j \omega t}##.

Now to find x

_{p}we must solve

$$x'' + \omega^2 x = \delta(t) = \left\{\begin{matrix} 0 \ \ if \ \ t \neq 0\\ \infty \ \ if \ \ t=0 \end{matrix}\right.$$

How can we find the particular solution to this equation?

Is this the correct approach so far? How does the given initial condition factor into the calculations?

Any help is greatly appreciated.