# 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.

• i think you will Need a numerical method – Dr. Sonnhard Graubner Sep 16 '17 at 18:11
• 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 – C.C.12 Sep 16 '17 at 18:19
• 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... – Raffaele Sep 16 '17 at 18:43
• unfortunately, but perhaps it is the only way – C.C.12 Sep 16 '17 at 19:30
• How many terms of $y$ do you want? – Nosrati Sep 16 '17 at 20:19

To simplify the Taylor series arithmetic introduce $u=\cos y$, $v=\sin y$. Then $$u'=-vy'\\ v'=uy'\\ y''+axu=0.$$ 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}

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 $u=\ln v$

$\dfrac{du}{dx}=\dfrac{1}{v}\dfrac{dv}{dx}$

$\dfrac{d^2u}{dx^2}=\dfrac{d}{dx}\left(\dfrac{1}{v}\dfrac{dv}{dx}\right)=\dfrac{1}{v}\dfrac{d^2v}{dx^2}-\dfrac{1}{v^2}\left(\dfrac{dv}{dx}\right)^2$

$\therefore\dfrac{1}{v}\dfrac{d^2v}{dx^2}-\dfrac{1}{v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{iaxv}{2}-\dfrac{iax}{2v}=0$

$2v\dfrac{d^2v}{dx^2}-2\left(\dfrac{dv}{dx}\right)^2-iaxv^3-iaxv=0$