second order ODE via variation of parameters

$$A'(x)u_1(x) + B'(x)u_2(x) = 0$$
why we desire A=A(x) and B=B(x) to be of this form? What is the basis that this form is valid?

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This is closely tied to the method of osculating parameters.
Suppose we wish to represent, with constant coefficients, some arbitrary function $u(x)$ with two linearly independent functions $u_1(x)$ and $u_2(x)$,
$$u(x) = A u_1(x) + B u_2(x).$$
In general this can not be done.
The best we can do is match the value of the function and its derivative at some point $x_0$,
$$\begin{eqnarray*} u(x_0) &=& A u_1(x_0) + B u_2(x_0) \\ u'(x_0) &=& A u_1′(x_0) + B u_2′(x_0). \end{eqnarray*}$$
The conditions above determine the osculating parameters, the constants $A$ and $B$.
$A$ and $B$ will be different depending on the point $x_0$.
In general this fit will be poor at points far from $x_0$.

The method of variation of parameters involves finding the osculating parameters $A$ and $B$ at every point.
That is, we let $A$ and $B$ be functions of $x$.
The condition that they are the osculating parameters is that they satisfy
$$\begin{eqnarray*} u_G(x) &=& A(x) u_1(x) + B(x) u_2(x) \\ u_G'(x) &=& A(x) u_1′(x) + B(x) u_2′(x), \end{eqnarray*}$$
just as above.
For the second equation to hold it must be the case that
$$A'(x)u_1(x) + B'(x)u_2(x) = 0.$$

Here’s the assumption. Simply to make the first derivative easier to deal with we are going to assume that whatever $u_1(t)$ and $u_2(t)$ are they will satisfy the following […]