A few queries of the method of variation of parameters I've been reviewing my knowledge on the technique of variation of parameters to solve differential equations and have a couple of queries that I'd like to clear up (particularly for 2nd order inhomogeneous ODEs), if possible.
The first is that, given the complementary solution, $y_{c}(x)=c_{1}y_{1}(x)+c_{2}y_{2}(x)$, to some 2nd-order inhomogeneous ODE: $$a_{2}(x)y''(x)+a_{1}(x)y'(x)+a_{0}(x)y(x)=f(x)$$ we assume that the particular solution $y_{p}(x)$ has the form $$y_{p}(x)=u_{1}(x)y_{1}(x)+u_{2}(x)y_{2}(x)$$ where $u_{1}(x)$ and $u_{2}(x)$ are arbitrary functions. 
Is the motivation for this ansatz that, upon taking derivatives of it and inserting the ansatz back into the original ODE, we find that the result includes the homogeneous solution plus some additional terms that we hope to be able to use to find a particular solution to the original ODE?!
The second is that, starting from this ansatz we note that we require that $y_{p}$ is a solution to the inhomogeneous equation, and upon inserting this into the ODE (and doing a little algebra), this leaves us with the equation $$a_{2}\frac{d}{dx}\left[u'_{1}y_{1}+u'_{2}y_{2}\right]+a_{1}\left[u'_{1}y_{1}+u'_{2}y_{2}\right]+ a_{2}\left[u'_{1}y'_{1}+u'_{2}y'_{2}\right]= f(x)$$ which applies a single constraint on the forms of $u_{1}(x)$ and $u_{2}(x)$. However, this would, in general, provide us with an infinity of particular solutions (as at least one of the functions, $u_{1}(x)$ or $u_{2}(x)$, remains completely arbitrary). We note, though, that we only require one particular solution and as such we are free to apply a further constraint of our choosing, in order to determine explicit forms for $u_{1}(x)$ and $u_{2}(x)$  [I'm a bit unsure whether my argument is correct here?!]. As such, we have one constraint (that the LHS equals the RHS [which has a fixed form $f(x)$]), and this leaves us with one degree of freedom that we are free to constrain. Thus, we choose that $$u'_{1}(x)y_{1}(x)+u'_{2}(x)y_{2}(x)=0$$ such that $$\left[u'_{1}(x)y'_{1}(x)+u'_{2}(x)y'_{2}(x)\right]= \frac{f(x)}{a_{2}(x)}$$ Is this the correct reasoning?
 A: Answers to your specific questions:

The first is that...Is the motivation for this ansatz that... the 
  result includes the homogeneous solution plus some additional terms 
  that we hope to be able to use to find a particular solution to 
  the original ODE?!

If you include arbitrary constants of integration, then yes, you'll get both the complementary solution and the particular solution all in one. Normally though, the constants of integration are left out and we get the particular solution to the non-homogeneous equation only and add the complementary solution to it afterward.

The second is that... However, this would, in general, 
  provide us with an infinity of particular solutions...

It provides you with only one particular solution, but an infinity of ways to get there.
You could instead choose $u_1(x)=0$ or that $u_2(x)=x\cos(x)e^{-x^2}$ or that $u_1(x)+u_2(x)=\sin(x)$, however, that could make finding the solution easier or harder. It depends on the integrals you end up evaluating. Depending on the specific problem, there might be a more clever choice to use up your degree of freedom besides $u_1'(x)y_1(x)+u_2'(x)y_2(x)=0$. I'll provide an example.
An Interesting Example:
Consider the equation $y''-2y'+y=e^t\sec^2(t)$. The complementary solution is $y_c=c_1e^t+c_2te^t$, and we'll let $y_1=e^t$ and $y_2=te^t$. The standard variation of parameters technique gives a particular solution of
$$\displaystyle Y_p(t)=-e^t\int t\sec^2(t)dt+te^t\int \sec^2(t)dt.$$
However, letting $u_2(t)=0$ instead simplifies to the equation
$$u_1''(t)=\sec^2(t)$$
This can be solved by integrating twice:
$$\displaystyle u_1(t)=\int  \left(\int \sec^2(t) \ dt\right) \ dt 
=\int  \tan(t) \ dt = \ln|\sec(t)|.$$
Thus $Y_p=e^t\ln|\sec(t)|$. This is slightly easier than the standard variation of parameters method, at least for this example. If instead we chose $u_1=0$ and solved for $u_2$, we end up with integrals that are more difficult.
