Solution of non-linear differential equation $y''+ay'+b\sin(y)\cos(y)=c$, where $a,b,c$ are constants. I am working on a project, in that project model I arrived at a differential equation. For further analysis, I need the solution of this equation. I tried to solve the equation but couldn't get success. Please anyone give some idea to solve this.
$y''+ay'+b\sin(y)\cos(y)=c$, where $a,b,c$ are the constants means they are combinations of parameter values of real life materials.
If $y'=0$ then it can be solved by multiplying the equation by $y'$ and integrating it. But, in the present case I'm not getting any clue. Thanks in advance.
Note: Ideas in the direction of approximated solution are also invited.
 A: $$y''+ay'+b\sin(y)\cos(y)=c$$
This ODE is  of autonomous kind. Thus the order can be reduced, that is to say transforming it to a first order ODE.
Change of variables :
$$\begin{cases}
Y=\frac{dy}{dx}\\
X=y
\end{cases}
\quad\implies\quad \frac{d^2y}{dx^2}=\frac{dY}{dX}\frac{dX}{dx}=\frac{dY}{dX}\frac{dy}{dx}=\frac{dY}{dX}Y$$
$$\boxed{Y\frac{dY}{dX}+aY+b\sin(X)\cos(X)=c}$$
This is the first order nonlinear ODE to be solved for $Y(X)$.
But the problem is far to be solved because this is an Abel's differential equation of the second kind.
Of course the solution exists but in general cannot be expressed with a finite number of standard functions. That is to say, up to now no special function has been defined and standardized in order to analytically express the solutions on a formal manner in the general case.
Conclusion : Don't expect a definitive answer. Better try to approximately solve your problem in terms of series expansion or thanks to numerical methods.
A: Here is a series expansion assuming the function has a Taylor Series:
$$y’’=-ay’-\frac b2 \sin(2y)+c=\implies y=\sum_{n=0}^\infty \frac{y^{(n)}(k)(x-k)^n}{n!}$$
Let $y(k)=c_0,y’(k)=c_1$ then,
$$y’’(k)=-ay’(k)-\frac b2\sin(2y(k))+c=-ac_1-\frac b2\sin(2c_1)+c$$
$$y’’’(k)=-a y’’(k)-by’(k)\cos(2y(k))= -a^2c_1+\frac {ab}2\sin(2c_1)-ac -bc_1\cos(2c_0)$$
$$y^{(4)}(k)= a^3c_1-\frac {a^2b}2\sin(2c_1)+a^2c +abc_1\cos(2c_0)+abc_1+\frac {b^2}2\sin(2c_1)-bc+2bc_1^2\sin(2c_0)$$
$\vdots$
There is a complicated pattern. Continue the process to find the series expansion of the general solution.
A: On scaling variables $t=as$, $x(t)=2y(s)$ and renaming constants $\beta=b/a^2$ , $\alpha=2c/a^2$, the equation reads
$$\tag{1}
x''+x'+\beta \sin x=\alpha
$$
This corresponds to the `damped pendulum with constant tangential force'. We can learn a lot about the equation without an exact solution. First, let us study the phase portrait. Let $v(t)=\frac{dx}{dt}$ then we can plot the vector field $(x',v')$ in the $x$-$v$ plane.

The image above shows the vector field (black) and an example trajectory (blue). On the left $\alpha=0.7$, $\beta=2.8$ and on the right $\alpha=1.2$, $\beta=0.5$. These are the two qualitatively different regimes, depending on which of $\alpha$ or $\beta$ is larger. The vector field is $2\pi$ periodic in the $x$ direction.
An equilibrium solution (left image), $x(t)=x_0=\text{constant}$, $x'=0=x''$ exists when
$$
\sin x_0=\frac{\alpha}{\beta}
$$
If $\alpha<\beta$, there are two equilibrium points. If $\alpha=\beta$, there is one, and if $\alpha>\beta$ there are none. Any such points are $2\pi$ periodic. If an equilibrium solution exists, then deviations from equilibrium $x(t)=x_0+x_1(t)$ are governed by the approximate linear differential equation (by substituting $x(t)=x_0+x_1(t)$into (1) and formally expanding everything in small $x_1$)
$$\tag{2}
x_1''+x_1'+x_1\cos x_0=0
$$
This is the damped oscillator equation. Note that $x_0$ is a known constant. The solutions are exponentials, and may be found by substituting $x_1(t)=e^{rt}$. When you do this, you will learn about the stability of the equilibrium points. I think that when they are two equilibria, one is stable and the other is not, based on the sign of $\cos x_0$. Furthermore, the stable point is attractive and the system always flows towards it (modulo the $2\pi$ periodicity). You'll need to treat the case $\alpha=\beta$ separately, I find that the single equilibrium point is still attractive in this case.
When $\alpha>\beta$, the system flows to an approximate steady state (right image), in which $x'$ is of bounded variation. To determine this, study the solution of (1) when $x'' \ll x'$, that is
$$\tag{3}
x' \sim \alpha-\beta\sin x \qquad,\qquad t\to\infty 
$$
This equation may be solved exactly, the result is (the integration constant disappears from the expression for large $t$)
$$\tag{4}
x(t)=2\tan^{-1}\left[\frac{\beta}{\alpha}-\sqrt{1-(\beta/\alpha)^2}\tan\left(\frac{\alpha t}{2}\sqrt{1-(\beta/\alpha)^2}\right) \right]
$$
Differentiating (4) to find explicitly $x'$, then finding the maxima and minima of $x'$ in the limit of large $t$ we find
$$\tag{5}
\alpha-\beta<v_\infty<\alpha+\beta
$$
Where $v_\infty$ is the large time asymptotic velocity. The red lines in the image on the right are the bounds $\alpha \pm \beta$.
In conclusion
The large $t$ behavior of the system is either to approach an attractive fixed point
$$
x(t)=x_0+x_1(t) \qquad,\qquad \alpha\leq \beta
$$
Where $x_1$ varies like $e^{-t}$, so the approach is exponential. Or, if $\alpha>\beta$, the system approaches the periodic motion described by (4) with the velocity described by (5). These behaviors are all independent of the initial conditions.
