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

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?

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: $$y=\left\{\begin{matrix} \pm x, & x \neq 1\\ 1, & x=1\\ \end{matrix}\right.$$

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 $y^2=cx^2$.

• Note that a piecewise defined functions don't has to be continuous or differentiable in general, but a solution of an ode is a differentiable function. This is a definition. – Fakemistake Jun 20 '17 at 11:42
• Note that we require differentiability on a closed interval containing the initial $x$ value for precisely the reason you mentioned: uniqueness will fail otherwise. In your "Side note" example, differentiability is only satisfied on the open set $\mathbb R\setminus\{1\}$, on which $y(x)=-x$. – user254433 Jun 23 '17 at 6:45
• You seem to confuse in the last edit the definition of a solution, which is always a continuously differentiable function, and the maximal information that you can extract from the differential equation with symbolic means. Most ODE do not even have that, there is simply no transformation that allows to integrate all terms of the equation. Textbook examples are really carefully selected for solvability. – LutzL Jun 23 '17 at 6:56

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 $y(0)=0$!)

• Thanks for the answer. I thought that general solutions can be relations, see my edit. – Abraham Zhang Jun 23 '17 at 6:32
• @EpicGuy I'm not sure what you mean by 'relation' here - sometimes relation is used as an expression for a class of functions or the like. But to satisfy an ODE, the solution needs to have a well-defined value and a well-defined derivative - otherwise, how would you check that $y'(x)=y(x)/x$? – Toffomat Jun 23 '17 at 7:17
• By "relation" I was trying to say that $y$ wasn't a function of $x$, such as the $y$ in $y^2+x^2=5$. So if I understand correctly, $y$ really means $y(x)$, and the relations provided by my textbook aren't relations at all, but, say for $y^2+x^2=5$, describing 2 functions at once? – Abraham Zhang Jun 23 '17 at 7:34
• Well, these are slightly different notions: $a^2 + b^2=7$ provides a relation between two numbers, $a$ and $b$. You can use that relation to express one of the numbers as a function of the other one, e.g. $a(b)=\sqrt{7-b^2}$ (but only within limitations - you have to be careful about domains of definition and the sign of the square root). On the other hand, a differential equation involve a function and its derivatives - if $y$ were not a function, what would $y'$ be? (The variable is often implicit, but always there.)... – Toffomat Jun 23 '17 at 10:27
• ... So in that sense, an ODE such as $y'(x)=y(x)/x$ relation between the (sought-after) function $y(x)$, its derivative and the independent variable. Note that the notation $y'=y/x$ is somewhat unclear - possibly, both $y$ and $x$ could be functions of some other variable, or $x$ could be a constant. It just customary in many circumstances that $x$ is the independent (real) variable. – Toffomat Jun 23 '17 at 10:31