Method of variation of parameters In the method of Variation of Parameters of solving differential equations, where do the values used for $y_1$, and $y_2$ come from? Are they roots of the homogenous equation? Also, I assume that I first need to standardize my given equation before I can retrieve my $g(x)$. Is this correct? 
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where do the values used for $y_1$, and $y_2$ come from? Are they roots of the homogenous equation?

Sorta, but the terminology "roots" is not accurate. For linear differential equation:
$$
y''(x) + p(x) y'(x) + q(x)y(x) = g(x),\tag{1}
$$
the $y_1$ and $y_2$ must form a fundamental set of solution for the homogeneous counterpart of $(1)$:
$$
y''(x) + p(x) y'(x) + q(x)y(x) = 0.
$$
So that the Wronskian $W$ of $y_1$ and $y_2$ is not zero at least on some open intervals of interest.


Also, I assume that I first need to standardize my given equation before I can retrieve my $g(x)$. Is this correct?

Correct, if I assume the term "standardize" means "make the coefficient of $y''$ be $1$". For example for the equation
$$
2x^2 y'' + 3x y' - y = g(x),
$$
the first step is to dividing $2x^2$ on each term:
$$
y'' + \frac{3}{2x} y' - \frac{1}{2x^2}y = \frac{1}{2x^2}g(x)\equiv h(x),
$$ 
so that you can use the variation of parameter formula:
$$
y_p = -y_1\int \frac{y_2 h}{W}dx + y_2\int \frac{y_1 h}{W}dx
$$
to find the particular solution. This "standardize" step you said also can tells us some extra information of the open interval where the solution exists for an initial value problem: the coefficients are discontinuous at $x=0$, so any interval on which a unique solution exists must not contain this point.
