How can I prove that for ordinary differential equations (ODE) the general solution involves *any* particular solution? I was struck by this sentence here:

Now, back to the work at hand. Notice in the last example that we kept saying “a” particular solution, not “the” particular solution. This is because there are other possibilities out there for the particular solution we’ve just managed to find one of them. Any of them will work when it comes to writing down the general solution to the differential equation.

It makes sense then that whenever possible one tries to resort to functions that differentiate (or second differentiate into themselves, plus or minus a sign or a constant), such as $e^x$ or $\sin(x),$ but how can I prove that any particular solution will do?
I am looking for an answer with connections to linear algebra, null space, etc.
 A: The set of all solutions of a linear non-homogeneous equation has a structure: it is a linear manifold in the space of all smooth functions. It is just as a straight line gives all solutions to an algebraic equation of the form $ax+by=c$. In the latter example, you can get all possible solutions (points on that line) by adding to ANY particular solution all the solutions of the corresponding homogeneous equation $ax+by=0$ (which is a parallel line through the origin). For linear differential equations, you solve the homogeneous equation first, getting linear combinations of "basic solutions" (exp, or sines, etc) and then you just add that linear space to any particular solution that you find (by inspection, by trying solutions in some form, etc.) It does not matter what particular solution you choose, you get the space of all solutions as a result. (compare adding a line through the origin to a point on the plane).
Half of any course on Linear Algebra is contained in the picture below. As you see, $p$ can be any point on the line, you still get the blue line by adding $p$ to the black line (linear subspace).

A: The idea is that a solution to the initial value problem $y''+py'+qy=g;\ y(x_0)=y_0,\  y'(x_0)=y'_0$ is uniquely determined by any particular solution, added to the general solution to the homogeneous case. More precisely, 
if $y_1$ and $y_2$ are linearly independent solutions to the homogeneous equation $y''+py'+qy=0$, then $\textit{every}$ solution to the initial value problem $y''+py'+qy=0;\ y(x_0)=y_0,\  y'(x_0)=y'_0$ has the form $y_h=c_1y_1+c_2y_2$ for suitable constants $c_1$ and $c_2$.
Now, suppose we have to solve the $\textit{non}$-homgeneous problem $y''+py'+qy=g;\ y(x_0)=y_0,\  y'(x_0)=y'_0$ where $g\neq 0.$ And suppose we can find a particular solution, $y$. Then, if $\tilde y$ is $\textit{any}$ other particular solution, $\tilde y-y$ is a solution to the $\textit{homogeneous}$ case, so it has the form $\tilde y-y=c_1y_1+c_2y_2,\ $ which implies that $\tilde y=y+c_1y_1+c_2y_2$.
We conclude from this that once we find a particular solution, any other one is uniquely determined. 
