Time period of ODE

Is it possible to find time period of the following non-linear ODE?

$$\frac{1}{\cos{y}}\frac{\mathrm{d}^2 y}{\mathrm{d} x^2 } = a \sin{y} + b, \quad y =y(x).$$

If so, how to obtain it? Is there a numerical method to approach this problem?

• You mean, $\frac{1}{\cos{y} }y'' = a \sin{y} + b$, right? Does this ODE have any initial / boundary conditions? What have you tried? Has this problem any physical background/motivation? Commented Aug 28, 2014 at 13:53
• @Demoreno Thanks.When b=0, special case of Sine-Gordon. Commented Aug 28, 2014 at 17:17
• I am calculating still.The time period answer given below by Robert Israel is relatable as arc length of a loop on fixed pseudo-sphere Gauss curvature and geodesic curvature of loop through parameters a to b respectively. Commented Aug 28, 2014 at 17:41
• @ Robert Israel: I posted this on SE Mathematica site as "Help to numerically solve ODE". Please take a look. Regards Commented Dec 21, 2014 at 21:31

The "energy" $E = (y')^2 - a \sin^2 y - 2 b \sin y$ is conserved. I'll suppose $a > 0$ and the initial conditions pick out a value of $E$ where there are two solutions $s_1 < s_2$ to $E = -a s^2 - 2 b s$ with $s$ in the interval $(-1,1)$. Note that $-a s^2 - 2 b s > E$ for $s_1 < s < s_2$. Then this solution will oscillate with period
$$2 \int_{\arcsin(s_1)}^{\arcsin(s_2)} \dfrac{dy}{\sqrt{-a \sin^2 y - 2 b \sin y - E}}$$
$$y'y''=y'(a\sin y \cos y +b\cos y)$$ Integrating, we obtain: $$y'^2=-\frac{a}{2}\cos(2y)+b\sin y+C=a\sin^2 y+b\sin y-\frac{a}{2}+C$$
For small y: $$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2 } = a y + b=a(y-(-b/a))$$ Recognizing SHM equation with amplitude $y_m$ $$\ddot y=\omega^2(y-y_m),T=2\pi/\omega$$ Now $$\omega=\sqrt a,T=2\pi/\sqrt a,|y_{m}|=b/a$$