# Why does the interval where the differential equation makes sense not include $x = 0$?

Here is the ODE equation $xy' = 2y$. A book states that the interval where it makes sense doesn't include $0$. Can you explain why? As to me, I think it is defined everywhere.

This problem is from the book "Ordinary Differential Equations - Harry Pollard, Morris Tenenbaum". The exact problem states: "Prove that the functions in the right-hand column below are solutions of the differential equations in the left-hand columns. (Be sure to state the common interval for which solution and differential equation make sense.)"

$xy' = 2y$ and $y = x^2$ with the answer $x \neq 0$

• $y'$ might have a discontinuity at $x=0.$ But otherwise you can define a $y$ that fits the diff eq. – Doug M Jun 28 '18 at 19:54
• @DougM, how it might have a discontinuity? – Turkhan Badalov Jun 28 '18 at 19:56
• The equation makes sense for $x=0$. It says that $y(x)=0$. If your interpretation of what the book is saying is accurate, then the book is wrong. A function that satisfies that equation for every $x$, must be of the form $y=Cx^2$ for $x\neq0$ and by continuity $y(0)=0$. – user569098 Jun 28 '18 at 19:56
• But check whether what they are saying is that or just excluding $x=0$ to then apply a method (say separation of variables, or integrating factor) to find the solution on $x\neq0$. Being the method, not the equation, the one that doesn't make sense for $x=0$. – user569098 Jun 28 '18 at 19:59
• I suspect your book intends to manipulate the equation into $\frac{y^\prime}{y} = \frac{2}{x}$ before solving, which is why they are (artificially) restricting $x$ to be nonzero. – Michael Biro Jun 28 '18 at 20:00

As equation, as LB_O commented, the formula makes sense. However, the equation does not allow to determine a slope for $x=0$ for $y\ne 0$, and for $y=0$ the slope is arbitrary. In that view the equation is not a differential equation at $x=0$.

In general the domain of an ODE $y'=f(x,y)$ is an open set where $f$ is continuous. For an implicit ODE $0=F(x,y,y')$ one demands that $0=F(x,y,v)$ has at least one solution $v$ for any $(x,y)$ in the domain and that $\partial_vF(x,y,v)$ is invertible there, or that at least a unique continuous solution $v=f(x,y)$ exists locally around that point.

If the determination of the domain of the ODE is focused on the $x$-axis as in the given task, then the domain in question has the form $I\times \Bbb R^n$ where $I$ is an open interval. Thus if one point $(x,y)$ is not in the domain, then the whole line $\{x\}\times\Bbb R^n$ is not in the domain.

Note that if you look at continuous continuations of solutions of the given ODE in $x=0$, then the general solution is $$y(x)=\begin{cases}C_1x^2&\text{ for }x\ge 0,\\C_2x^2&\text{ for }x<0,\end{cases}$$ with independent constants $C_1,C_2$.