General solutions to differential equations, are we missing valid relations?

Question

Our highschool maths class has just begun solving simple ODE's, and have been introduced the idea of using a constant to represent a family of functions (relations?). However, I can frequently find many more solutions than the answer provided.

For example, consider $y' = \frac y x$. The standard solution is $y=cx$, where in every relation within the family, $c$ takes only one value. Can't $c$ take on multiple values for each particular solution in the solution family? The resulting graphs of $y$ against $x$ will have many curves, nonetheless they will still satisfy the ODE as it's implicit.

Open intervals are another way to generate more types of solutions. In the case of $y' = \frac y x$, the derivative of a solution mustn't exist when $x=0$, so then can't $y$ equal anything at that point?. Would it be more complete for $c$ to take different values depending on whether $x$ is positive or negative? (This may result from the antiderivative of $\frac 1x$ discussed in this answer.) If this was the case $y = |x|$ can be a solution. Following from this, we can give many values to $y$ for open intervals of $x$, such as $$y=\left\{\begin{matrix} c_1x, & a_1< x< b_1\\ c_2x, & a_2< x< b_2\\ \vdots\\ c_nx, & a_n< x< b_n\\ \end{matrix}\right.$$ where choices for $a$, $b$, and $c$ ensure that every possible solution within the Cartesian plane is included, and, with open intervals, that there are no endpoints where the derivative is undefined. Why don't we consider this situation in the general solution?

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Side note: Some questions treat $y$ like a function, give an initial condition (i.e. $y=1$ when $x=1$), and you are asked to find the particular solution. If we consider the ODE from before, $y=x$ is the expected relation. But similarly, I can construct other values for $y$ that keeps the relation differentiable for the domain of the ODE ($x\in\Bbb{R}\setminus\{0\}$), doesn't conflict with the initial condition, and still satisfies the ODE: $$y=\left\{\begin{matrix} \pm x, & x \neq 1\\ 1, & x=1\\ \end{matrix}\right.$$

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Edit: It was pointed out that this is avoidable if we limit the general solutions to functions. I'm not sure about this because the textbook that our class is working on often has relations as the general solution to ODE's. If needed, a simple solution that's not a piecewise relation is $y^2=cx^2$.

Answer 1

You seem to be on a right track, but to mix up some issues. I'll try to discuss that following your example ODE.

In $y'=y/x$ your're looking for a function $y(x)$ satisfying that equation. In particular, for a given value of $x$, there has to be exactly one value $y(x)$ (in moregeneral situations, you have to worry about domains of definition, but that's not the main pooint here), so your parameter $c$ cannot 'take on multiple values at the same time' -- if it did, what would be the value of, say, $y(1)$? On the other hand, if you change the constant $c$ into a function of $x$, even a piecewise-constant one, you change the derivative and your ODE is not satisfied.

Furthermore, your function has to be differentiable, because otherwise the ODE is not well-defined, much less satisfied. Hence, a function $$y(x)=\left\{\begin{matrix} -x &x\neq 1\\1&x=1\end{matrix}\right.$$ is not a good solution - it has a jump at $x=1$, so it is not differentiable there.

However, what you seem to be in the process of finding out is that an ODE can (and generically does) have multiple solutions, and you might wonder whether you found all of them. For the theory you can look at https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem, but the general point is that a first-order ODE (meaning that it contains only first derivatives) needs one 'initial condition' to uniquely specify the solution. For your case, there is a family of solutions, distinguished by a single parameter $c$ (and that family indeed contains all solutions). Hence, the value of, e.g., $y(1)$ is required to pick one out of the family of solutions, and your constant $c$ will be just $y(1)$. (Note that the derivative of $y$ is perfectly well-defined at $x=0$ as long as $y(0)=0$!)