# The general solution of $y''=-\sin y$

When I asked Mathematica to solve the ODE $$y''=-\sin y \tag{1}$$ I got the solutions $$y=\pm 2 \text{am}\left(\frac{1}{2} \sqrt{\left(c_1+2\right) \left(t+c_2\right){}^2}|\frac{4}{c_1+2}\right), \tag{2}$$ where $\text{am}(u|m)$ is the Jacobi Amplitude function. I wonder why there is a $\pm$ ambiguity here (I believe it's related to the square root), since the equation $(1)$ is explicit.

P.S.

Using the rules $c_1 \mapsto -2+4c_1^2,c_2 \mapsto c_2/c_1$ one gets the equivalent form $$y= \pm2 \text{am}\left(\sqrt{\left(t c_1+c_2\right){}^2}|\frac{1}{c_1^2}\right)$$ and if one cancels the square root with the square, noticing that am is odd in the first argument one gets an even nicer form $$y=\pm2 \text{am}\left(c_1t +c_2|\frac{1}{c_1^2}\right).$$ However, my question still remains: Why is there an ugly $\pm$ in the solution if the ODE is written explicitly in the form $y''=f(x,y,y')$? Is there a nicer form for the general solution?

Thank you!

• $y''$ and $\sin y$ are both odd functions of $y$, so if some particular function $y(t)$ is a solution then $-y(t)$ must also be a solution. – Qiaochu Yuan Jan 18 '17 at 4:57
• @QiaochuYuan That's true. However, we have a similar situation with $y''=-y$ and this time the solution can be expressed unambiguously as $y=c_1 \cos(t)+c_2 \sin(t)$. – user1337 Jan 18 '17 at 5:04
• That's not unambiguous; there are a whole two continuous parameters there! In any case, that differential equation is linear, so its space of solutions forms a vector space, which is very much not the case here. – Qiaochu Yuan Jan 18 '17 at 5:06
• Simple harmonic motion of a pendulum happens when the angle is small enough to use the first term of the Taylor series for sine, beyond that the effects of large amplitude on the period become apparent. – Triatticus Jan 18 '17 at 6:18

$$y=\pm2 \text{am}\left(c_1t +c_2\bigg|\frac{1}{c_1^2}\right).$$ The Jacobi amplitude function is symmetrical : $\quad\text{am}(-x|k)=-\text{am}(x|k)$ $$2\text{am}\left(c_1t +c_2\bigg|\frac{1}{c_1^2}\right)=-2\text{am}\left(-(c_1t +c_2)\bigg|\frac{1}{c_1^2}\right) = -2\text{am}\left(C_1t +C_2\bigg|\frac{1}{C_1^2}\right)$$ with $C_1=-c_1$ and $C_2=-c_2 .$

This means that, considering the general solution of the ODE one can forget the $\pm\:$: $$y=\pm2 \text{am}\left(c_1t +c_2\bigg|\frac{1}{c_1^2}\right) \equiv 2\text{am}\left(c_1t +c_2\bigg|\frac{1}{c_1^2}\right) \text{insofar } c_1,c_2 \text{ are any constants.}$$ Of course, this isn't true if we don't consider the whole set of solutions, but one particular solution, according to initial/boundary conditions : The conditions determine the constants and the sign. That is why it is better to write : $$y=\pm2 \text{am}\left(c_1t +c_2\bigg|\frac{1}{c_1^2}\right)\:,$$ knowing that, for the particular solution, $\pm$ means $+$ or $-$ (i.e.: only one solution, not both).

NOTE :

The question itself is somehow ambiguous : " I wonder why there is a $\pm$ ambiguity here , since the equation $(1)$ is explicit ".

It isn't specified if the subject of the question concerns an ODE alone, or an ODE with initial/boundary conditions.

Not only in the case of equation $(1)$, but in all cases of ODEs : If the question concerns an ODE without initial/boundary conditions "explicit" doesn't mean that only one solution exist. They are an infinity of solutions since the constants such as $c_1$, $c_2$, (and if they are $\pm$ in it), all can be chosen among many values. One cannot say that there is an ambiguity due to the presence of arbitrary constants and $\pm$ in the general solution.

If the question concerns an ODE with initial/boundary conditions, insofar the conditions are consistent for a unique solution, those conditions determines the values of the constants and the signe affected to $\pm$.