Solutions to the differential equation $ax''+bx=c$. I have the following differential equation:
$$a\cdot x''(t) + b\cdot x(t)=c$$
where $a$, $b$, and $c$ are positive real numbers.
The associate polynomial would be:
$$ s^2+b/a=0$$
thus,
$$s=\pm j \sqrt{\frac{b}{a}}= j \omega_{0} $$
It is well know that the solution of this differential equation would be in the form:
$$ x(t)=A_{0} \cdot \sin (\omega_{0} t) + A_{1}\cdot \cos (\omega_{0} t) + A_{2} $$
where $A_{0}$, $A_{1}$ and $A_{2}$ are real numbers.
The second derivative is
$$x''(t)=-\omega_{0}^2 \cdot [A_{0} \cdot \sin (\omega_{0} t) + A_{1}\cdot \cos (\omega_{0} t)]$$
However, if I replace this solution on the differential equation something interesting is happenning:
$$ a\cdot x''(t)+b \cdot x(t)= (b-a\cdot \omega_{0}^2 )\cdot [A_{0} \cdot \sin (\omega_{0} t) + A_{1}\cdot \cos (\omega_{0} t)] + b \cdot A_{2}=c$$
The term $b-a\cdot \omega_{0}^2$ is always zero. So the constants $A_{0}$ and $A_{1}$ are undefined for any initial condition. There are infinite solutions for this differential equation. What I'm doing or assuming wrong when I get this kind of statement?
 A: Your equation becomes $A_2=\frac{c}{b}$, which means that the set of all solutions to your differential equation is the set
$$\{A_0\sin(\omega_0t) + A_1\cos(\omega_0t)| A_0, A_1\in\mathbb R\}$$
Indeed, for every pair of real numbers $A_0, A_1$, the function $A_0\sin(\omega_0t) + A_1\cos(\omega_0t)$ solves the differential equation. There are infinitely many solutions.

To actually calculate a single solution, you also need the initial conditions, for example, $x(0)$ and $x'(0)$, before you can calculate it.
A: What you wrote is correct as a way to find the value of $A_2$ from the equation (this is the non-homogeneous part of the equation, a constant).
The constants $A_0$ and $A_1$ remains arbitrary as they define the general solution of the homogeneous equation. (As the equation is of the second order, there is a double infinity of solutions.)
Now to exploit the initial condition, you write
$$ x_0=x(0)=A_{0} \cdot \sin (\omega_{0} 0) + A_{1}\cdot \cos (\omega_{0} 0) + A_{2} $$ and $$A_1=x_0-A_2.$$
$A_0$ requires an additional condition, such as the initial derivative.
