Laplace transform of Diff. Eq containing sine Function

[dimensionless]

Homework Statement

Find the Laplace transform of the pendulum equation.

Homework Equations

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

$$s = \sigma + i \omega$$

The Attempt at a Solution

Taking the laplace transform I get

$$s^{2} + \alpha s + g \frac{\omega}{s^{2}+\omega^{2}} = 0$$

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

$$s^{2} + \alpha s + g \frac{\omega}{-\omega^{2}+\omega^{2}} = 0$$

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
$$\mathcal{L} \left\{\sin(t)\right\} = \frac{1}{s^{2}+1}$$

which would make the transform

$$s^{2} + \alpha s + g \frac{1}{s^{2}+1} = 0$$

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:

$$\frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin( \theta ) = f(t)$$

With the Y(s) terms I get:

$$s^{2} Y(s) + \alpha s Y(s) + g \frac{1}{s^{2}+1}} = F(s)$$

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

$$\frac{Y(s)}{F(s)}$$

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