# 2nd order linear ODEs: why does $y''$ have no coefficient in the standard form?

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.

• 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. Jul 18, 2023 at 9:24
• @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. Jul 18, 2023 at 9:47