# Sinusoidal Diff EQ

1. Jun 21, 2008

### SonOfOle

1. The problem statement, all variables and given/known data
$$\ddot{\Theta}=C \sin{\Theta}$$ where $$\Theta$$ is a function of time, and C is a constant.

I ran into this on a Lagrangian Equation problem, and though the problem doesn't ask for the solution, I'm wondering how one would solve this Diff EQ. I'm afraid my intro to Diff EQ class was a couple years back and I'm coming up short with any ideas.

Thanks.

2. Jun 21, 2008

### EngageEngage

what function do you differentiate twice to get a sin with a constant in front?

3. Jun 21, 2008

### Defennder

You can solve this by noting that the second order derivative of theta is purely a function of theta alone. This means that $$\theta'' = f(\theta)$$

By the chain rule:

$$\frac{d^2 \theta}{dt^2} = \frac{d}{d\theta} \left( \frac{{\theta'}^2}{2} \right) = C \sin \theta$$

This is separable and solvable. Once you have solved for $\theta '$, you can use solve for theta again using separation of variables

4. Jul 5, 2008

### SonOfOle

Alright, so:

$$\frac{d^2 \theta}{dt^2} = \frac{d}{d\theta} \left( \frac{{\theta'}^2}{2} \right) = C \sin \theta$$

$$\left( \frac{{\theta'}^2}{2} \right) = (C_{1} \cos \theta) + C_{2}$$

$$\theta = \int \sqrt{(C_{1} \cos \theta)+C_{2}} dt$$

How would I use separation of variables here?

5. Jul 5, 2008

### HallsofIvy

Staff Emeritus
By separating the variables before you integrate!
$$\theta '= \sqrt{2C cos(\theta)+ C_2}$$
Where I have incorporated the "2C2[/sup]" into the unknown constant C2 (but not C1= 2C; C is a given number, not an unknown).

$$d\theta= \sqrt{2C cos(\theta)+ C_2}dt$$
$$\int \frac{d\theta}{\sqrt{2C cos(\theta)+ C2}}= \int dt[/itex] That integral on the left looks to me like an "elliptic integral" which cannot be integrated in terms of elementary functions. 6. Jul 5, 2008 ### Defennder That's not what you should get. You should have [tex] \frac{1}{\sqrt{c_1 - 2 \cos \theta}} d\theta = dt$$

But I see a problem here. I don't know of a way to integrate the LHS.

7. Jul 5, 2008

### SonOfOle

Thanks all. That helps. It's coming back, just slowly at times.

8. Jul 5, 2008

### nicksauce

Of course, the traditional way to solve it is to make a small angle approximation, and forget about the cases where that isn't valid. :p