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.

  • $\begingroup$ i think you will Need a numerical method $\endgroup$ Sep 16, 2017 at 18:11
  • $\begingroup$ Unfortunately I need a non-numeric solution. As I already wrote, it would be very good to have a series solution, and I believe that in series of powers can be found, but I have not managed to retrieve it after several attempts $\endgroup$
    – C.C.12
    Sep 16, 2017 at 18:19
  • $\begingroup$ A series solution with $\cos y$ is a mess! $$\cos y(x)=\frac{1}{6} x^3 \left(-y^{(3)}(0) \sin (y(0))+y'(0)^3 \sin (y(0))-3 y'(0) y''(0) \cos (y(0))\right)+x^2 \left(-\frac{1}{2} y''(0) \sin (y(0))-\frac{1}{2} y'(0)^2 \cos (y(0))\right)-x y'(0) \sin (y(0))+\cos (y(0))+O(x^4)$$ I have only written the first $4$ terms... $\endgroup$
    – Raffaele
    Sep 16, 2017 at 18:43
  • $\begingroup$ unfortunately, but perhaps it is the only way $\endgroup$
    – C.C.12
    Sep 16, 2017 at 19:30
  • $\begingroup$ How many terms of $y$ do you want? $\endgroup$
    – Nosrati
    Sep 16, 2017 at 20:19

2 Answers 2



Let $y=iu$ ,

Then $\dfrac{dy}{dx}=i\dfrac{du}{dx}$


$\therefore i\dfrac{d^2u}{dx^2}+ax\cos iu=0$

$\dfrac{d^2u}{dx^2}-iax\cosh u=0$


Let $v=e^u$ ,

Then $u=\ln v$






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

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