Laplace transform of Diff. Eq containing sine Function

[dimensionless]

Homework Statement

Find the Laplace transform of the pendulum equation.

Homework Equations

The pendulum equation:
<br /> \frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin(\theta) = 0 <br />

<br /> s = \sigma + i \omega<br />

The Attempt at a Solution

Taking the laplace transform I get

<br /> s^{2} + \alpha s + g \frac{\omega}{s^{2}+\omega^{2}} = 0<br />

I usually drop the \sigma term in s = \sigma + i \omega, which leaves

<br /> s^{2} + \alpha s + g \frac{\omega}{-\omega^{2}+\omega^{2}} = 0<br />

The \frac{\omega}{-\omega^{2}+\omega^{2}} is problematic, but I'm not sure what to do. I don't see the usefulness of the \sigma variable, and I'm not sure that I'm doing this correctly to begin with

 

[Feldoh]

You're not doing the transform correctly. For laplace you're just taking the real part of z in exp^{-zt}, z = \sigma + i \omega.

 

[dimensionless]

I'm not sure what you're saying but I had a thought. I think the Laplace of the sine function is
<br /> \mathcal{L} \left\{\sin(t)\right\} = \frac{1}{s^{2}+1}<br />

which would make the transform

<br /> s^{2} + \alpha s + g \frac{1}{s^{2}+1} = 0<br />

This is more sensible, but I still need to verify it some how.

 

[Feldoh]

What do you get if you take the Laplace transform of just a second deriavative?

 

[dimensionless]

I think it is something like

\mathcal{L} \left\{y(t)\right\} = s^{2} Y(s)

presuming that
\dot y(0) = 0
and
\ddot y(0) = 0

I think the sine term would make this a nonlinear differential equation, and I'm now I'm thinking that maybe the Laplace transform doesn't work on nonlinear differential equations.

 

[Feldoh]

Sure it does!

Why do you keep dropping the Y(s) terms? That's what you need to take the inverse laplace transform of! Solve for Y(s) and take the inverse transform.

 

[dimensionless]

I'm trying to get a transfer function of:

<br /> \frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin( \theta ) = f(t) <br />

With the Y(s) terms I get:

<br /> s^{2} Y(s) + \alpha s Y(s) + g \frac{1}{s^{2}+1}} = F(s)<br />

For the transfer function, I want to Y(s) and F(s) on the same side like this:

<br /> \frac{Y(s)}{F(s)}<br />

but I don't see and way to factor things out that way.