# Differential equation [closed]

I am trying to solve the following differential equation;

$$\frac{d^2 x}{d t^2}=-\omega^2 x \delta(t-t^\prime).$$

I know this is of the form

$$x(t)= A \sin(\omega t) + B \cos(\omega t).$$

However this delta Dirac function is confusing me. My reasoning is the following; An increase in acceleration at $$t^\prime$$ results in an increase of constant velocity starting from $$t^\prime$$ and a linear constant increase in distance so;

$$x(t) = \left[A \sin(\omega t) + B \cos(\omega t)\right] t \Theta(t>t^\prime).$$

With $$\Theta$$ being the step function. This still feels more like a guess than a straightforward answer, then I am not 100% sure of the first derivative.

Any help would be welcome.

• This question may be better suited for the Mathematics SE. – Davide Morgante Aug 26 at 16:21
• Comment to the post (v3): Are you sure the delta function is multiplied rather than added on the rhs? – Qmechanic Aug 26 at 16:58
• @Qmechanic You know I was just thinking the same thing. The equation as stated is not dimensionally consistent. – Philip Aug 26 at 17:15
• Okay, you could be right.. The lagrangian that is given is the following: $L=\frac{m \frac{dx}{dt}^2}{2} - \frac{m \omega^2 x^2}{2} \delta (t-t')$. With the Euler Lagrange equations I arrived at the diff eq.. (part of the exercice) – Alex Aug 26 at 17:25
• @Alex The Lagrangian isn't dimensionally consistent either! The delta function here has dimensions 1/time. The second term does not have dimensions of energy. Furthermore, your problem is not an oscillator, it looks more like a free particle before and after $t'$, since $\delta(t-t')=0$ for $t\neq t'$. I'm assuming that there's a mistake in the original source. – Philip Sep 1 at 8:02

You need to find the solutions for $$t> t'$$ and for $$t> t'$$ and then impose the integration constants by the boundary conditions. The problem is actually very similar to that of a delta-function barrier in quantum mechanics, but with zero energy.