# Solving differential equation with a step impulse.

1. Oct 31, 2013

### Fionn00

Solving differential with a step impulse.

Hi,
I have problem I know I should be able to do but I've been stuck on it for a while. Just looking to be pointed in the right direction.

(dq^2/d^2t) + 2*ζ*ω*dq/dt + (ω^2)*q = u(t)/L

Where u(t) is a step impulse, q is the charge through an inductor. So q and dq/dt (the current) are both = 0 at t = 0.

So I have to solve q(t) for the damping constant ζ <1, >1 and =0.

I tried getting the laplace of it and then getting Q(t) on one side but solving the inverse laplace of the result from tables is impossible and I'm not sure that's how this is supposed to be solved as I've seen many different solutions to similar problems.

Is Laplace the right thing to do? And if so what am I missing to solve it.
If not Laplace then what? Should I be plugging in an e^-st for q or something?

Thanks any help is appreciated.

2. Oct 31, 2013

### vanhees71

The Laplace transform should of course work. The back transformation is usually done using complex integration an the theorem of residues etc.

A direct way is to solve the equation directly. You first need the full solution of the homogeneous equation, i.e., with vanishing right-hand side. For $\zeta \neq 1$ a simple exponential ansatz leads to two linearly independent solutions, and for $\zeta=1$ you get one exponential solution and another one of the form $t \exp(-\omega t)$.

Finally you need one particular solution of the inhomogeneous equation, and for $t>0$ the right-hand side is simply a constant. Thus the ansatz $q(t)=\text{const}$ gives you such a solution. Now you put everything together and use the freedom from the two parameters in the superposition of the two independent solutions of the homogeneous equation to fix your boundary conditions.

3. Oct 31, 2013

### Staff: Mentor

What is a step impulse? I've heard of a unit step and a unit impulse, but not a step impulse. If you meant a unit impulse, then there should not be an s in the first brackets. To invert the transform, you need to complete the square in the second brackets; or use the quadratic formula to factor the second brackets, and then separate into partial fractions.

Chet

4. Nov 1, 2013

### vela

Staff Emeritus
Apparently, you're analyzing an RLC circuit with the three elements in series. q is the charge on the capacitor, not the inductor, and dq/dt is therefore the current flowing through the capacitor. Now because the elements are in series, dq/dt is also the current that flows through the resistor and inductor.

Laplace is fine, but you need to learn how to invert the transform. What did you get for Q(s)? Use partial fractions to break it up into terms with linear or quadratic denominators.

5. Nov 1, 2013

### rude man

To help you choose the best transform pairs it would be helpful if you could scan or take a picture of your Laplace transform pair table.