Is this differe equation (and its solution) already known? While studying for an exam, I met the following nonlinear differential equation
$a\ddot{x}+b\dot{x}+c\sin x +d\cos x=k$
where $a,b,c,d,k$ are all real constants. My teacher says that this differential equation does not admit closed form solution, but on this I would like to compare myself with you. Is this equation (and its solution) already known?
Thank you very much

$a\ddot{x}+b\dot{x}+c\sin x+d\cos x=k$
$a\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+c\sin x+d\cos x-k=0$
This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0317.pdf
Let $\dfrac{dx}{dt}=u$ ,
Then $\dfrac{d^2x}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dx}\dfrac{dx}{dt}=u\dfrac{du}{dx}$
$\therefore au\dfrac{du}{dx}+bu+c\sin x+d\cos x-k=0$
$au\dfrac{du}{dx}=-bu-c\sin x-d\cos x+k$
$u\dfrac{du}{dx}=-\dfrac{bu}{a}-\dfrac{c\sin x+d\cos x-k}{a}$
This belongs to an Abel equation of the second kind.
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $u=\dfrac{1}{v}$,
Then $\dfrac{du}{dx}=-\dfrac{1}{v^2}\dfrac{dv}{dx}$
$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dx}=-\dfrac{b}{av}-\dfrac{c\sin x+d\cos x-k}{a}$
$\dfrac{dv}{dx}=\dfrac{(c\sin x+d\cos x-k)v^3}{a}+\dfrac{bv^2}{a}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
