Unsolvable first order nonlinear differential equation? $\frac{dx(t)}{dt} + \sin(x(t)) = \sin(\omega t)$ I do want to solve the following differential equation analytically:
$$\frac{dx(t)}{dt} + \sin(x(t)) = \sin(\omega t)$$
I tried several methods to solve this equation, unfortunately without any success. In the last week, I have read a lot of papers related to that kind of prototype and have looked into all the books that deal with first order differential equations. It still seems to be impossible. Neither Wolfram-alpha, nor Matlab's symbolic toolbox can give me a solution for it. Because of that, I am wondering if there is actually a solution for that kind of differential equation?
Thanks for your help! 
 A: Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $\omega=1$

You can plot just a fundamental patch $t\in[0,2\pi]$, $x\in[-\pi,\pi]$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution):

Alternative visualization:

Edit. Code for the plots, in Mathematica:
sol = ParametricNDSolve[{
    x'[t] + Sin[x[t]] == Sin[t],
    x[0] == x0},
   x, {t, 0, 4 \[Pi]}, {{x0, -3 \[Pi], 3 \[Pi]}}];
Plot[Evaluate@Table[
   x[x0][t] /. sol, {x0, -3 \[Pi], 2.5 \[Pi], .25}], {t, 0, 3.5 \[Pi]}]

per = NDSolve[{
    x'[t] + Sin[x[t]] == Sin[t],
    x[0] == x[2 \[Pi]]},
   x, {t, 0, 2 \[Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 \[Pi]}, {x, -\[Pi], \[Pi]}],
 Plot[Evaluate[x[t] /. per], {t, 0, 2 \[Pi]}, PlotStyle -> Red]]

A: Surprisingly, this differential equation can be "solved analytically", albeit in terms of non elementary functions.
First we have the equation
$$\frac{dx(t)}{dt} + \sin(x(t)) = \sin(\omega t)$$
which i will rewrite as
$$\frac{dx}{dt} + \sin(x) = \sin(\omega t)$$
(to make it easier type)
now substitute
$$x= -\frac{\cos(\omega t)}{\omega} + u$$
therefore $$\frac{dx}{dt}= \sin(\omega t) + \frac{du}{dt}$$
after rearranging the equation becomes
$$\frac{du}{dt} + \sin\left(-\frac{\cos(\omega t)}{\omega} + u\right) =0$$
which can be rewritten as
$$\frac{du}{dt} = \sin\left(\frac{\cos(\omega t)}{\omega} - u\right)$$
now using the identity $\sin(\theta) = \frac {e^{i\theta}-e^{-i\theta}}{2i}$
our equation can be rewritten as
$$\frac{du}{dt} = \frac{e^{i\left(\frac{\cos(\omega t)}{\omega} - u\right)}-e^{i\left(\frac{-\cos(\omega t)}{\omega} + u\right)}}{2i}$$
now we rearrange the terms to arrive at
$$2i\frac{du}{dt} = \frac{e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}}{e^{iu}}- \frac {e^{iu}} {e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}}$$
now substitute $e^{iu} = v$
therefore $$ \frac{dv}{dt}=iv \frac{du}{dt}$$
Also $$ \frac{2}{v} \frac{dv}{dt} = \frac{e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}}{v}- \frac {v} {e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}}$$
Now we can use variable separation method by grouping the terms having $v$
in the denominator and then rearrange the equation as
$$ \frac{1}{v^2} \frac{dv}{dt} = \frac{1}{{e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}}{\left(e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}-2\right)}}$$
for now we shall focus on the RHS (right hand side)
$$ RHS = \frac{-1}{2{e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}}} + \frac{1}{2{\left(e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}-2\right)}} $$
(By method of partial fractions)
Now to solve this ode we must integrate this equation with respect to t
NOTE: These integrals are non elementary
To avoid clutter i will integrate each term of the RHS separately
let $$I_1 =\frac{-1}{2} \int e^{-i\left(\frac{\cos(\omega t)}{\omega}\right)}dt$$
Substitute $p=-i\left(\frac{\cos(\omega t)}{\omega}\right)$
Also $dp=i \sin(\omega t)dt=i(1+(w^2p^2))^{\frac{1}{2}}dt $
Thus
$$I_1 =\frac{-1}{2} \int \frac{e^{p}}{i(1+(w^2p^2))^{\frac{1}{2}}}dp$$
Taylor series expansion of$$ \frac{1}{(1+(w^2p^2))^{\frac{1}{2}}}= \sum_{n=0}^{\infty} \frac{(2n-1)!!\,(-1)^n (w)^{2n} p^{2n}}{(2^n) (n!)}$$
Now substitute this in the integral, separate the constants and interchange the
integral and summation
$$I_1 =\frac{i}{2} \sum_{n=0}^{\infty} \frac{(2n-1)!!\,(-1)^n (w)^{2n} \int{e^p p^{2n}}}{(2^n) (n!)}dp$$
Finally $$I_1 =\frac{i}{2} \sum_{n=0}^{\infty} \frac{(2n-1)!!\,(-1)^n (w)^{2n}\: \Gamma(2n+1,-p)}{(2^n) (n!)}$$
Where $\Gamma$ is the incomplete gamma function and $p=-i\left(\frac{\cos(\omega t)}{\omega}\right)$
Now let $$I_2 =\frac{1}{2} \int \frac{1}{\left(e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}-2\right)}dt$$
Substitute $z=e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}$ in the expansion of $\frac{1}{z-2}$
To obtain $$\frac{1}{e^{i\left(\frac{\cos(\omega t)}{\omega}\right)}-2}=\sum_{n=0}^{\infty} (-1)\left(\frac{1}{2}\right)^{1+n}\:e^{ni\left(\frac{\cos(\omega t)}{\omega}\right)}$$
Now substitute the summation in the integral and separate its first term
$$I_2 =\frac{-1}{4} \int {\left(1 + \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n}\:e^{ni\left(\frac{\cos(\omega t)}{\omega}\right)}\right)}dt$$
Interchange the summation and integral sign and simplify to obtain
$$I_2 =\frac{-1}{4} t + \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n}\:\int e^{ni\left(\frac{\cos(\omega t)}{\omega}\right)}dt$$
Now let $I_3=\int e^{ni\left(\frac{\cos(\omega t)}{\omega}\right)}dt$
Similar to how $I_1$ was 'solved', $I_3$ can be 'solved'
by substituting
$$Q_n=ni\left(\frac{\cos(\omega t)}{\omega}\right)$$
So $I_2$ will evaluate to a double summation
The LHS will evaluate to a function of x
(The final solution is too long to type in neatly)
