Explicit solution of parametric solutions of an ODE

Question

I need to find the explicit solution of the following ODE:

$y'+\sin y'=x$, $y=y(x)$.

I have found these two parametric solutions:

$x=t+\sin t$ and $y=\frac{t^2}{2}+t\sin t+\cos t+c$, $c\in\Bbb R$.

Any idea if it exists?

Edit: My teacher mentions in the rubric of the exercise to find the solutions in closed form. My question is if closed-form solutions include parametric solutions.

Answer 1

let $f$ be defined by $$f(y) = y + \sin y.$$ it is not hard to see that $f$ is invertible. we have $$ f(y') = x \to y' = f^{-1}(x)$$ integrating the last equation gives $$y = \int f^{-1}(x) \, dx \tag 1$$ we can make a change of variabe $u = f^{-1}(x), x = f(u), dx = f'(u) \, du = (1 + \cos u)du.$ with this change of variable $(1)$ becomes $$ y = \int u (1 + \cos u) \, du = \frac12 u^2+u\sin u + \cos u + c, u = f^{-1}(x).$$