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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

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What is the independent variable of this ODE? – doraemonpaul Oct 28 '12 at 20:00
Is t.Indeed x is a function of t: $x=x(t)$ – Mark Oct 28 '12 at 20:10
I think too many cases should be divided. Does there any constants have some restrictions for not equal to some values? – doraemonpaul Oct 29 '12 at 0:03
$a,b,c,k>0$, $d<0$. These are the only limitations. Sorry if I have not written before. – Mark Oct 29 '12 at 6:30
up vote 1 down vote accepted

$a\ddot{x}+b\dot{x}+c\sin x+d\cos x=k$

$a\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+c\sin x+d\cos x-k=0$

This belongs to an ODE of the form

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

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