Solve for $x(t) : x(t)= b+at+ ∫ ( ∫\sin (wt-kx(t)) \, dt) \, dt$ $$ x(t)= b+at+\int \left (\int \sin (wt-kx(t)) \,dt \right ) dt$$
How do I extract $x(t)$ from the above equation. Any hint? Or is this equation has been already solved?
Thanks in advance for any help!
 A: $$ x(t)= b+at+\int \left (\int \sin (wt-kx(t)) dt \right ) dt \tag 1$$
With $x(t)=X(t)+b+at$
$$ X(t)= \int \left (\int \sin\left(wt-kX(t)-kb-kat\right) dt \right ) dt$$
Differentiate two times :
$$\frac{d^2X}{dt^2}=\sin (wt-kX(t)-kb-kat) $$
Let $y(t)=wt-kX(t)-kb-kat$
$ X=\frac{1}{k}(wt-y-kb-kat)\quad;\quad X'=\frac{1}{k}(w-y'-ka)\quad;\quad X''=-\frac{y''}{k}$
$$-\frac{y''}{k}=\sin(y)$$
$$2y''y'=-2k\sin(y)y'\quad\to\quad (y')^2=2k\cos(y)+c_1$$
$$y'=\pm\sqrt{2k\cos(y)+c_1}$$
This is a separable ODE. The solution $x(y)$ is an elliptic integral. The inverse function is a Jacobi elliptic function :
$$y(t)=2\text{ am}\left(\frac{\sqrt{(2k+c_1)(t+c_2)}}{2}\:\Bigg|\:\frac{4k}{2k+c_1} \right)$$
am$(\:|\:)$ is the Jacobi Amplitude elliptic function : http://mathworld.wolfram.com/JacobiAmplitude.html
$$x(t)=\frac{1}{k}\left(wt-2\text{ am}\left(\frac{\sqrt{(2k+c_1)(t+c_2)}}{2}\:\Bigg|\:\frac{4k}{2k+c_1} \right)\right)+kb+kat$$
If the integrals in Eq.$(1)$ are defined, one have to bing back the above result into Eq.$(1)$ in order to determine the constants $c_1$ and $c_2$.
