How solve this nonlinear trigonometric differential equation sHello to everyone, i would like a suggestion on how solve this nonlinear differential equations:
$$y''+ a\ x\cos y=0 $$
where $a \in \mathbb{R}\ and \ \ y=y(x) $
I am aware that if a solution can be found it will be a series solution,
for example, proceeding with the power series could lead to some result?
Or what replacement do I need to bring it back to the first order?
Thank you in advance for any idea or proposed solution.
P.S. I have not yet been able to find out if this differential equation has already been resolved in literature.
 A: Hint:
Let $y=iu$ ,
Then $\dfrac{dy}{dx}=i\dfrac{du}{dx}$
$\dfrac{d^2y}{dx^2}=i\dfrac{d^2u}{dx^2}$
$\therefore i\dfrac{d^2u}{dx^2}+ax\cos iu=0$
$\dfrac{d^2u}{dx^2}-iax\cosh u=0$
$\dfrac{d^2u}{dx^2}-\dfrac{iaxe^u}{2}-\dfrac{iaxe^{-u}}{2}=0$
Let $v=e^u$ ,
Then $u=\ln v$
$\dfrac{du}{dx}=\dfrac{1}{v}\dfrac{dv}{dx}$
$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(\dfrac{1}{v}\dfrac{dv}{dx}\right)=\dfrac{1}{v}\dfrac{d^2v}{dx^2}-\dfrac{1}{v^2}\left(\dfrac{dv}{dx}\right)^2$
$\therefore\dfrac{1}{v}\dfrac{d^2v}{dx^2}-\dfrac{1}{v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{iaxv}{2}-\dfrac{iax}{2v}=0$
$2v\dfrac{d^2v}{dx^2}-2\left(\dfrac{dv}{dx}\right)^2-iaxv^3-iaxv=0$
A: To compute a power series up to any desired order, you can use the tools of Taylor series arithmetic that are a part of the Automatic/Algorithmic Differentiation theory. As differentiation of a power series contains an index shift, new coefficients result on the left side while evaluating or updating the evaluation on the right side.
To simplify the Taylor series arithmetic introduce $u=\cos y$, $v=\sin y$. Then
\begin{align}
u'&=-vy'\\
v'&=~~~uy'\\
y''&=-axu.
\end{align}
This is now a problem where the multiplication of Taylor series is the most complicated operation which allows to establish the equations for the coefficients of these 3 series
$$y=\sum_{n=0}^\infty y_kx^k, ~~~u=\sum_{n=0}^\infty u_kx^k,~~~  v=\sum_{n=0}^\infty v_kx^k.
$$
These can then be computed by successively inserting the already computed coefficients,
\begin{align}
y_0&=y(0)&  u_0&=\cos(y_0),&  v_0&=\sin(y_0),\\
y_1&=y'(0)&  u_1&=-v_0y_1,&    v_1&=u_0y_1,\\
2y_2&=0,&  2u_2&=-v_0(2y_2)-v_1y_1,&  2v_2&=u_0(2y_2)+u_1y_1,\\
6y_3&=-au_0,&    3u_3&=-\sum_{k=1}^3v_{3-k}(ky_k),&    3v_3&=\sum_{k=1}^3u_{3-k}(ky_k),\\
&\vdots\\
n(n-1)y_n&=-au_{n-3},&nu_n&=-\sum_{k=1}^nv_{n-k}(ky_k),&    nv_n&=\sum_{k=1}^nu_{n-k}(ky_k),\\
\end{align}
For further information on the automatic creation of such algorithms, see

*

*FADBAD++ by Stauning and Bendtsen (project active 1996-2003), the third example in the crash-course section of their description site,

*VNODE, used for examples in papers by N. Nedialkov (2001)

*TSM - ODE playground by user m4r35n357 (active 2018-), Taylor series manipulation with a focus on ODE

