# How to solve or approximate the result of the following differential equation?

Where $C_0, C_1, C_2, C_3, C_4$, and $\mu$ are constants, I want to solve or approximate the solution of the following differential equation.

\begin{align} & C_0 \cos \theta + C_1 \sin \theta \\ & - \ddot{\theta}(\mu C_2 \cos 2\theta + C_2 \sin 2\theta - \dfrac{\mu}{2} C_3 \sin 2\theta + \dfrac{\mu}{2} C_4 \sin 2\theta + C_3 \cos ^2 \theta + C_4 \sin ^2 \theta) \\ & - {\dot{\theta}}^2 (C_2 \cos 2\theta - \mu C_2 \sin 2 \theta - \dfrac{C_3}{2} \sin 2\theta + \dfrac{C_4}{2} \sin 2\theta - \mu C_3 \cos ^2 \theta - \mu C_4 \sin ^2 \theta) \\ &= 0 \end{align} where $$\ddot{\theta} = \dfrac{d^2 \theta}{dt^2}$$ $$\dot{\theta} = \dfrac{d \theta}{dt}$$

I have spent a week trying to solve this problem yet I cannot seem to be able to find a way to approach this differential equation. I will be grateful for any kind of help\advice given. The conditions are $\theta(0) = \dot{\theta}(0) = 0$

• What do you mean with "approximate"? You can find numerical approximations for any set of constants with the usual means. – Lutz Lehmann Dec 26 '17 at 10:42
• Lutzl I am sorry for being unclear. I did mean numerical approximations(as far as I am aware of) yet I do not know how to do it in the "usual means". May I ask how? – Isamu Isozaki Dec 26 '17 at 12:13
• Transform into a first order system and feed it into the ODE solver of your choice. Fixed-step Runge-Kutta methods are the most basic and easiest to implement. – Lutz Lehmann Dec 26 '17 at 12:35

The equation has the general form

$$f(\theta) - \ddot{\theta}g(\theta) - {\dot{\theta}}^2 h(\theta) = 0$$

where \begin{align} f(\theta) &= C_0\cos\theta + C_1 \sin\theta \\ g(\theta) &= \left(\mu C_2 + \frac{C_3}{2} - \frac{C_4}{2}\right)\cos2\theta + \left(C_2 - \frac{\mu C_3}{2} + \frac{\mu C_4}{2}\right)\sin 2\theta + \frac{C_3}{2} + \frac{C_4}{2} \\ h(\theta) &= \left(C_2 - \frac{\mu C_3}{2} + \frac{\mu C_4}{2} \right)\cos 2\theta - \left(\mu C_2 + \frac{C_3}{2} - \frac{C_4}{2} \right)\sin 2\theta - \frac{\mu C_3}{2} - \frac{\mu C_4}{2} \end{align}

Using the general method for an autonomous system of second order, let $\omega = \dot{\theta}$. Then

$$\omega \frac{d\omega}{d\theta}g(\theta) + \omega^2 h(\theta) = f(\theta)$$

Observe that $$h(\theta) = \frac12 g'(\theta) - \frac{\mu}{2} (C_3 + C_4)$$

If $\frac{\mu}{2} (C_3 + C_4) = 0$, then $h(\theta) = \frac12 g'(\theta)$ and the equation is equivalent to

$$\frac{d}{d\theta}\left(\frac{\omega^2}{2} g(\theta)\right) = f(\theta)$$

Then we have an analytic solution in terms of $$\omega(\theta) = \pm\sqrt{\frac{2}{g(\theta)}\int_0^\theta f(\phi) d\phi}$$

and the inverse function is given by $$t(\theta) = \int_0^\theta \frac{1}{\omega(\phi)} d\phi$$

As for a numerical approximation, someone else will have to give a better answer than me.

The general trick is to transform the original equation into a first-order system \left\{ \begin{aligned} \dot{\theta} &= \omega \\ \dot{\omega} &= \dfrac{f(\theta) - \omega^2 h(\theta)}{g(\theta)} = F(\theta,\omega) \end{aligned} \right.

The most basic approach is to use Euler's method, which applies a simple linear interpolation for the next time step \left\{ \begin{aligned} \theta(t + \Delta t) &= \theta(t) + \dot{\theta}(t)\Delta t = \theta(t) + \omega(t)\Delta t \\ \omega(t + \Delta t) &= \omega(t) + \dot{\omega}(t)\Delta t = \omega(t) + F(\theta,\omega)\Delta t \end{aligned} \right. where $\Delta t$ is some arbitary small step.

You are free to explore other methods with higher accuracy, such as the Runge-Kutta family.

Do note that this method fails when $g(\theta) = 0$. As in, your solution will be bounded in $\theta \in [0,\theta_0)$ for some $g(\theta_0)=0$.

• Thank you! I really do appreciate the help and the intuitive explanation so far! – Isamu Isozaki Dec 26 '17 at 12:22
• You can also use any numerical method to approximate the solution since the equation is in the form $\ddot{\theta} = f( \theta, \dot{\theta})$ – Dylan Dec 26 '17 at 12:30
• Wait, can you please send me some links or pdfs? I am quite new to the world of differential equations and am not quite up to par in its methods. – Isamu Isozaki Dec 26 '17 at 14:31
• By the way, (please forgive my ignorance) am I correct in assuming $\frac{d}{d\theta}\left(\frac{\omega^2}{2} g(\theta)\right) = f(\theta)$ works because $\dot{\theta}\dfrac{d\dot{\theta}}{d\theta}g(\theta)$ evaluates to 0 because $\dfrac{d\dot{\theta}}{d\theta}$ evaluates to 0?? – Isamu Isozaki Dec 26 '17 at 14:40
• Generaral guide to numberical methods – Dylan Dec 27 '17 at 0:20