Method of Variation of Parameters - Assigning zero works? I have yet to find a decent answer on this, and so I don't think this question is inappropriate. Also, this question is mainly meant for people that are very familiar with this method.
In the method of variation of parameters method for solving the non-homogeneous (heterogeneous?) part of a second order linear differential equation - that is, one in the form $$y''(t)+a(t)y'(t)+b(t)y(t)=g(t)$$I'm still boggled with the exact purpose of setting parts of the first derivative of the particular solution equal to solution.
I will explain what I'm talking about here.
Suppose we have a $2^{nd}$ order linear non-homogeneous ODE,  $$y''(t)+a(t)y'(t)+b(t)y(t)=g(t)$$
Also, suppose we've already obtained a general solution for the homogeneous version of the ODE, $y''(t)+a(t)y'(t)+b(t)y(t)=0$, that's in the nice form $$y_h=c_1y_1+c_2y_2$$ To get the particular solution for the non-homogeneous part, we start off with the weird (but successful) method of variation of parameters - change the $c_1$ and $c_2$ to $u_1(t)$ and $u_2(t)$ respectively (hence, variation of parameters), and call that $y_p$.
$$y_p=u_1(t)y_1+u_2(t)y_2$$ 
The goal is to find out $u_1(t)$ and $u_2(t)$, and so we now get the $1^{st}$ and $2^{nd}$ derivatives of $y_p$ so we can slap them into the original ODE and hopefully solve for $u_1(t)$ and $u_2(t)$. The first derivative follows by chain rule:
$$y'_p = u'_1(t)y_1 + u_1(t)y'_1 + u'_2(t)y_2 + u_2(t)y'_2$$
Here is the part I don't understand.
In every textbook I've seen this in, the next step goes basically as follows:

For no reason and no motivation (as far as we're concerned right now), we now set the following:
  $$u'_1(t)y_1 + u'_2(t)y_2 = 0$$
  This will turn out to work in the end very nicely, and it will also make our computations much more simple.

I've seen full well that this does work out nicely, and that it does make the computations much more simple, but my question is why is this done, and why does it work out so nicely? Why does that equation $$u'_1(t)y_1 + u'_2(t)y_2 = 0$$ always have to be satisfied for particular solutions to $2^{nd}$ order linear non-homogeneous ODE's of this form?.
 A: The explanation that you are looking for relies on the fact that $y_1$ and $y_2$, being linearly independent solutions to the homogeneous equation, form a basis for the solution space of the inhomogeneous equation. This means that they span the space, so that the solution vector $\begin{pmatrix}y_p \\ y_p^{\prime}\end{pmatrix}$ can be written as a linear combination of the vectors $\begin{pmatrix}y_1 \\ y_1^{\prime}\end{pmatrix}$ and $\begin{pmatrix}y_2 \\ y_2^{\prime}\end{pmatrix}$. Hence, for some functions $u_1(t)$ and $u_2(t)$: $$\begin{pmatrix}y_p \\ y_p^{\prime}\end{pmatrix} = u_1\begin{pmatrix}y_1 \\ y_1^{\prime}\end{pmatrix} + u_2\begin{pmatrix}y_2 \\ y_2^{\prime}\end{pmatrix}$$
We consider the two components:
$$y_p = u_1y_1 + u_2y_2$$
$$y_p^{\prime} = u_1y_1^{\prime} + u_2y_2^{\prime}$$
By differentiating the first equation and subtracting it from the second we get that
$$0 = u_1^{\prime}y_1 + u_2^{\prime}y_2$$
which is what you were looking for. In my opinion, the subtlety lies in the observation that the above vectors form a basis for the solution space. Perhaps someone has a good explanation of why that is the case.
A: This is an extended comment on @whosleon answer.
I want to point out that to write $\begin{pmatrix}y_p(t) \\ y_p^{\prime}(t)\end{pmatrix} = u_1(t)\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix} + u_2(t)\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}$  with coefficients being functions $u_1(t)$  and $u_2(t)$ we just need to have two vectors $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}$ and $\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}$ be linearly independent for each $t$. This is true for $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}=\begin{pmatrix}y_1(t) \\ y_1^{\prime}(t)\end{pmatrix}$, $\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}=\begin{pmatrix}y_2(t) \\ y_2^{\prime}(t)\end{pmatrix}$  where $y_1(t)$ and $y_2(t)$ are independent solutions to any homogeneous linear second order ODE, no matter what specific equation we consider; it is also true for $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}=\begin{pmatrix}  1\\0\end{pmatrix}, \begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}=\begin{pmatrix}  0\\1\end{pmatrix}$. 
This does not explain why using the solutions of the associated homogeneous equation $\begin{pmatrix}w_1(t) \\ z_1(t)\end{pmatrix}=\begin{pmatrix}y_1(t) \\ y_1^{\prime}(t)\end{pmatrix}$, $\begin{pmatrix}w_2(t) \\ z_2(t)\end{pmatrix}=\begin{pmatrix}y_2(t) \\ y_2^{\prime}(t)\end{pmatrix}$ is a good choice, in the sense of giving  computable  $u_1(t)$ and $u_2(t)$.
My attempt at writing one possible explanation is in the answer to Intuition behind variation of parameters method for solving differential equations
Another potentially helpful idea is in an answer here: second order ODE via variation of parameters
