# 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?

• Why do you think we can guess what you're denoting by $y_1$, $y_2$ and $g(x)$? – joriki Jul 7 '11 at 21:14
• My guess would be that the system is $y''+a(x)y'+b(x)y=g(x)$ and $y_1$, $y_2$ are a fundamental set of solutions to the homogeneous equation $y''+a(x)y'+b(x)y=0$. But that is just my guess. – Matt Jul 7 '11 at 22:07
• Do you have some particular equation in mind? This is a lot simpler to respond if you share work on a problem so that there isn't any confusion as to notational differences. – davidlowryduda Jul 7 '11 at 23:56
• I second the suggestion that you write out the whole deal so we can see what you mean by the symbols you use. Meanwhile, you may be able to get enough information from the Wikipedia page en.wikipedia.org/wiki/Variation_of_parameters to get you over the hump. – Gerry Myerson Jul 8 '11 at 1:11
• I think that makes sense. Thanks! – user10695 Jul 8 '11 at 1:59

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.