Rephrase the question, as the first was erroneously written.
Hello everyone and thank you in advance for your attention,this is the nonlinear differential equations:
$$ y''+ a[\sin(y+b)]=f(x) $$
where $a,b \in \mathbb{R}$, $y=y(x)$ and $f:\mathbb{R} \to \mathbb{R}$.
I already solved the homogeneous equation (which it is a Lineard's non-linear differential equation ), but cannot apply the method of Lagrange (variation of parameters)as it is done with linear differential equations,how can i solve this non homogeneous equation?
Thank you in advance for any idea or proposed solution.
For completeness, and help your readers, i add that to solve the homogeneous equation $$ y''+ a[\sin(y+b)]=0 $$ i took this change of variable: $$y'=Z(y) \implies y''=Z'Z$$ but now it does not seem that such a change of variable can me solve the non-homogeneous equation,and i got stuck, unable to find the right change of variable or the right way to solve it.
It could bring to a Abel nonlinear differential equation?
$$ y''+ a[\sin(y)]+ b[\cos(y)] =f(x) \tag 1$$
$a[\sin(y)]+ b[\cos(y)]=\rho \sin(y+\phi)\quad\text{with}\quad \begin{cases}\rho=\sqrt{a^2+b^2}\\ \tan(\phi)=\frac{b}{a} \end{cases}$
Change of function :$\quad y(x)=u(x)-\phi\quad$ transform Eq.(1) into Eq.(2) : $$u''+\rho\sin(u)=f(x) \tag 2$$
Change of variable :$\quad t=\sqrt{\rho}\:x\quad\to\quad \frac{d^2u}{dt^2}+\sin(u)=\frac{1}{\rho}f\left( \frac{t}{\sqrt{\rho}}\right)$
Let $\quad F(t)=\frac{1}{\rho}f\left( \frac{t}{\sqrt{\rho}}\right).\quad$ Since $\rho$ and $f$ are known, $F(t)$ is a known function. $$\frac{d^2u}{dt^2}+\sin\left(u(t)\right)=F(t) \tag 3$$ Probably, no simpler form of equation (with no parameter inside) can be derived.
In the particular case $F(t)=0$ the solution can be expressed in terms of Jacobi elliptic function.
In the case $F(t)=C\neq 0$, Eq.(3) is an ODE of the autonomous kind. But the integration cannot be done in term of standard special functions.
A fortiori, in the general case $F(t)$ not constant, there is no closed form for the solution of the ODE.
This doesn't mean that closed form never exists in some particular cases. Just look backwards : Put a given function $y(x)$ into Eq.(1). This gives a particular function $f(x)$. For this function $f(x)$ , at least a closed form solution exist for Eq.(1) : the $y(x)$ a-priori chosen.
Thus, no definitive answer can be given to the question raised. This depends on the explicit definition of the function $f(x)$. But it is clear that analytical solving Eq.(1) is not possible in general.