Numerical integration of nonlinear second order equation I have (or, rather, a friend whom I'm trying to help has) a very messy differential equation, which I thought I'd try to solve numerically. However, I'm a little confounded as to what approach to use when doing this.
The equation is on the form
$\ddot\alpha(t) = f(\dot\alpha(t), \alpha(t), \dot\varphi(t), \varphi(t))$
where $\alpha(t)$ and $\varphi(t)$ are both unknown functions of time, to be determined (and $f$ is a crazy beast1...). I have initial values for both the functions and their first derivatives, so it should just be a matter of choosing an integrator and start to iterate.
However, all of the examples I find on e.g. Wikipedia are all for first-order equations, and I'm a little unsure what integrator to use. It's too long ago I took a basic course in numerical methods so even if they are applicable to this problem anyway, I don't remember how to adapt them :p
Is there a good integrator that works on a problem such as the one stated above? If so, pointers to some good explanation of it would be appretiated =)
Update:
Upon asking my friend for an expression for $\ddot\varphi$, in order to make a substitution and transform this into a first-order problem, it turned out that she had such an expression. However, when trying to transform the equations, I end up with a system on the form
$\dot\beta = f(\beta,\alpha,\gamma,\varphi,\dot\gamma)$
$\dot\gamma = g(\beta,\alpha,\gamma,\varphi,\dot\beta)$
$\dot\alpha = \beta$
$\dot\varphi = \gamma$
because, as you can see in the original problem2, the second order derivatives also depend on each other. How do I tackle this? Another substitution? Or can I somehow decouple these equations first, so the second order derivatives are independent of each other?

1The full problem, as it was originally stated for me:
$\displaystyle \ddot\alpha = \frac{(l\dot\alpha\cos(\alpha+\varphi)-g\sin\varphi)\sin(\alpha+\phi)+mlh\dot\varphi^2\cos(\alpha+\varphi)+V\cos\alpha}{E-Ml^2\sin^2(\alpha+\varphi)}$
 All quantities except $\alpha$ and $\varphi$, and their derivatives, are known constants.
2The full problem, but now stated in a different form:
$I\ddot\alpha = Mlh\ddot\varphi\sin(\alpha+\varphi)+Mlh\dot\varphi^2\cos(\alpha+\varphi)+V\cos\alpha$
$h\ddot\varphi = l\ddot\alpha\sin(\alpha+\varphi)+l\dot\alpha^2\cos(\alpha+\varphi)-g\sin\varphi$
 A: This problem wasn't so hard to solve, after all, as soon as I remembered how to classify it; it's implicit, which is the biggest part of the reason I struggled. This is how to solve it:


*

*Rewrite it as a first-degree problem, using substitutions as suggested by Arkamis. I had a system of two second-degree equations, so I ended up with four first-degree equations.
$
\left\{\begin{array}{rcl}
I\dot\alpha-Mlh\dot\beta\sin(\theta+\phi)-Mlh\beta^2\cos(\theta+\phi)-V\cos\theta  & = & 0 \\
h\dot\beta-l\dot\alpha\sin(\theta+\phi)-l\alpha^2\cos(\theta+\phi)+g\sin\phi & = & 0 \\
\dot\theta-\alpha & = & 0 \\
\dot\phi - \beta & = & 0
\end{array}\right.
$

*This system s on the form $f(t,y,\dot y)=0$, with $y=(\alpha,\beta,\theta,\phi)$, and can be solved with any implicit ODE solver, such as Matlab's ode15i. That's what I did =)
A: All second-order problems can be converted into a system of first order problems.
$$\ddot{y} = f(t,y) \longrightarrow y_1 = y,\ y_2 = \dot{y_1} \longrightarrow \dot{y_1} = y_2,\ \dot{y_2} = f(t,y_1)$$
A: If you're just looking for a numerical integration, you've done all the decoupling you need to.  First, you have to assign values for your initial $\beta$, $\alpha$, $\gamma$, and $\phi$, as well as your initial $\dot{\beta}$, $\dot{\alpha}$, $\dot{\gamma}$, and $\dot{\phi}$.  Note that initial $\dot{\beta}$, $\dot{\alpha}$, etc. cannot be derived from initial $\beta$, $\alpha$,etc.  You have to decide what they are.  Now you are ready to plug all of these into your favorite numerical integrator along with your formulae, and run it a time step to get your next set of values, and then plug them in again etc.  For numerical methods, you might Google the "leapfrog" method to start with, and once you're confortable with that, try 4th order Runge-Kutta.
