I'm studying second order linear ordinary differential equations.

I have learned that they take the form:

$y'' + p(x)y' + q(x)y = f(x)$

Why does $y''$ have no (or a constant $1$) coefficient here? What if, like $y'$ and $y$, it had an arbitrary $j(x)$ (not necessarily constant) coefficient? Does it affect anything at all?

I see that in order to give $y''$ a coefficient, let's call it $j(x)$, we can simply multiply the equation by $j(x)$. But that would mean that the coefficients of $y'$ and $y$ are necessarily divisible by $j(x)$. Which I guess narrows down the space of equations that answer to the form in question.

  • 2
    $\begingroup$ Because you can always divide throughout by the leading coefficient of $y''$ to normalize it. This is the same with polynomials. You can always normalize it such that the leading coefficient is 1. $\endgroup$
    – Nasser
    Jul 18, 2023 at 9:24
  • $\begingroup$ @Nasser I think I see where I was confused. The thing that was puzzling for me, was that after multiplying the equation by j(x), we get an equation where the coefficients of y' and y are necessarily divisible by j(x). Which seems to limit the space of conforming equations. But, I think that I was wrong in this understanding. Because when p(x) or q(x) are fractions, multiplying by j(x) will not necessarily create a number that is 'divisible' by j(x), because it won't necessarily be an integer. $\endgroup$
    – Aviv Cohn
    Jul 18, 2023 at 9:47


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