Particular solution of forced oscillation equation

Question

Consider the following ODE $$ \ddot{x}+2\dot x +2x=2l+l\cos t $$ for some constant $l$. This describes the movement of a forced oscillator. In order to find the general solution, I first find a particular solution, and then add it to the solution of the homogeneous equation. A hint, given in the original problem, is to consider solutions of the form $$ x(t)=A+B\cos t+C\sin t, $$ where $A,B$ and $C$ are constants.

What is the motivation behind such hint in this particular case? Why this form? Also, apart from looking at the general solution for any ODE of the form $$ \ddot{x} +p(t)\dot x+q(t) x=f(t), $$ is there an intuitive way of 'guessing' particular solutions, in cases where $f(t)$ is 'relatively' simple?

The highly ambiguous and informal quotation marks are because I feel this particular case is simply somewhat justifiable by the fact that $f(t)$ in the case above is defined on trigonometric functions, and derivatives of trigonometric-based solutions have periodic properties that allow for an easy guess. Any further insights are appreciated.

Answer 1

Yes. There is a systematic way to find particular solutions in case the right hand side (the force) in the equation is a linear combination of products of polynomials, sines/cosines or exponentials.

See https://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

In those cases you look for solutions of a specific form and you are guaranteed to find one and only one by solving a simple linear system.

Answer 2

If $x_1$ $x_2$ are the solution of the homogeneous ODE: $$x''+2x'+2x=0~~~(1)$$, theb the solutions of $$X''+2X'+2X=f(t)~~~~(2)$$ is given by $$X=C_1(t)~ x_1(t) + C_2(t)~ x_2(t)~~~(3)$$ where $$C_1(t)= -\int\frac{x_2(t) f(t)}{W(t)} dt+ D_1, ~~~ C_2(t)= \int \frac{x_1(t) f(t)}{W(t)} dt+D_2~~~~(4)$$ $D_1$ $D_2$ are constants and $W(x_1,x_2)$ is the Wronskian defined as $$W(x)=[ x_1 x_2'-x_1' x_2]~~~~(5)$$ Here in the case of (1) , we have $x_1=e^{-t} \sin t, x_2=e^{-t} \cos t, f(t)=2L+ L \cos t$. One can find the total solution (3) by finding $C_1(t)$ and $C_2 (t)$.