TITLE: Do you lose solutions when differentiating to solve an integral equation?
QUESTION [41 upvotes]: The question is more general but here's the problem that motivated it: I want to find all solutions to the integral equation
$$f(x) + \int_0^x (x-y)f(y)dy = x^3.$$
Differentiating twice with respect to x this yield the second order differential equation
$$f''(x) + f(x) = 6x.$$
The solution to this last equation is something of the form $f(x) = A \cos(x + \phi) + f_p$ where $f_p$ can be found by variation of parameters or other tools and $A, \phi$ are determined by initial values.
However, what I want to know is whether I'm accounting for all solutions here. Do I not "lose information" when differentiating the original equation? If so, how can I construct all solutions from the ones I just found?

REPLY [10 votes]: Something that I think that high school math doesn't explain very clearly is the theory of extra/missing solutions. Let's say we have equations $(1)$ and $(2)$. If $(1)\rightarrow (2)$, that means $(\text{x satisfies (1)})\rightarrow(\text{x satisfies (2)})$, so every solution of $(1)$ will also be a solution of $(2)$. However, $(1)\rightarrow (2)$ does not guarantee that every solution of $(2)$ will be a solution of $(1)$. So when you have a chain of equivalent equations, every solution of the final equation will be a solution of the original equation. When you have a chain of implication, you can have extraneous solutions. To lose solutions, you have to have a situation where $(2)\rightarrow (1)$, i.e. your later equation is sufficient, but not necessary, for the initial equation to be true. For instance, if you go from $x^2=y^2$ to $x=y$, the first equation is a consequence of the second, but the second does not necessarily follow from the first.
In your case, you are differentiating. If equation $(1)$ is $y=f(x)$, and equation $(2)$ is $y'=f'(x)$, we have $(1)\rightarrow(2)$ (if two functions are equal, then their derivatives are equal as well), but we don't have $(2)\rightarrow(1)$ (that is, if two derivates are equal, that doesn't mean the original functions were equal). So, for instance, if you have $f(x) = x+1$ and take the derivative of both sides, you get $f'(x) = 1$, and $f(x)=x$ is a solution to the second equation, but not the first.
Differentiating does in fact lose information, but losing information means that you are getting more solutions, not fewer. The more information you have, the more possibilities you can eliminate. If you know $y=x+1$, there are some ordered pairs that you can eliminate (for instance, you that (1,0) is not a solution). If you also know that $2y=3x$, then you can eliminate more ordered pairs. Remember that equations don't rule solutions in, they only rule solutions out.