Trigonometric Homogenous Differential Equation

I have the following nonlinear differential equation (I am using $y$ as shorthand $f(x)$):

$$\sin(y - y') = y''$$

I have tried the following

$$\cos(y - y')(y'-y'') = y'''$$ $$-\sin(y - y')(y'-y'')^2 + \cos(y - y')(y''-y''') = y''''$$ $$-y''(y'-y'')^2 + \dfrac{y'''}{y'-y''}(y''-y''') = y''''$$ $$-y''(y'-y'')^3 + y'''(y''-y''') = y''''(y'-y'')$$

But this looks pretty unhelpful. Is there a better way to solve this equation?

• Unless I misunderstand, it's not a functional equation, just a nonlinear differential equation. Commented Feb 27, 2015 at 21:54
• Sorry, I thought differential equations were a subset of functional equations. I'll edit that out.
– k_g
Commented Feb 27, 2015 at 21:55

i don't know how useful this is to you but here it is. we will make a change of variable $$y-y' = u.$$ then the differential equation $y'' = \sin (y-y')$ can be transformed into $$\sin u = y''= y'-u'=y-u-u'$$ now we have two first order equations

\begin{align}\frac{dy}{dx} &= y - u\\ \frac{du}{dx} &= y - u -\sin u\end{align}

the equilibrium solutions are $u = k \pi, v = k\pi$ are saddles with eigenvalues $\frac{-1 \pm \sqrt5}2$ for $k$ even and unstable spirals with eigenvalues $\frac{1 \pm \sqrt 3 i}2$ for $k$ odd.

• Wow. OK that's interesting. I originally found this equation while trying to generate interesting curves using differential equations, so the complexity is actually good news!
– k_g
Commented Feb 27, 2015 at 22:43

There's not much hope of closed-form solutions. You could use numerical methods or series.

• Could you provide a little more imformation? Like can the solutions be parametrified with two variables?
– k_g
Commented Feb 27, 2015 at 22:00
• That's always the case for a second order differential equation. You can use e.g. $u(0)$ and $u'(0)$ as the parameters. Commented Feb 27, 2015 at 23:39
• OK thanks; I wasn't sure.
– k_g
Commented Feb 28, 2015 at 0:12