# An first integral of nonlinear differential equation as like forced pendulum nonlinear diff. eq.

I'm trying to face this nonlinear differential equation:

$$y''(x)+\omega^2\sin\,y(x)=a\,x \,\;(1)$$

and I'm interested to found the solution of $$y'(x)$$ (an first integral)

The homogeneous part of previous one ode is like a nonlinear free (non forced) pendulum diff. eq. : $$\theta''+\omega^2\sin\theta=0$$ Then, the ode that i'm trying to solve it's similar to forced pendulum differential equations.

The first integral of homogeneous solution of (1) it's easy to solve : $$\frac{(y'(x))^2}{2}-\omega^2\cos \,y=\mathrm{const}$$

But, is there a solution for fist integral of (1) in a case of forcing of the type $$f(x)=a\,x$$ or otherwise?

• What do you consider the "trend" of $y'$ (or $y$)? Do you mean something like a linear function where the remainder is a periodic function and some small perturbation? – Lutz Lehmann May 17 '19 at 15:33
• No you are right, I had expressed myself badly.I want found y'(x). I corrected the text. – C.C.12 May 17 '19 at 15:37

For large $$x$$ the right side is large, while the $$\omega^2 \sin(y)$$ term is bounded. Thus it may be useful to consider this differential equation as a perturbation of $$y'' = ax$$. We can write $$y(x) = \sum_{k=0}^\infty \omega^{2k} y_k(x)$$ where \eqalign{y_0(x) &= y(0) + y'(0) x + a x^3/6\cr y_1(x) &= - \int_0^x dt \int_0^t ds\; \sin(y_0(s))\cr y_2(x) &= - \int_0^x dt \int_0^t ds\; y_1(s) \cos(y_0(s))\cr \text{etc}}
I suspect that each $$y_k$$ for $$k \ge 1$$ will be asymptotic to some straight line as $$x \to \infty$$.