I'm working through Exercise $3$ in Tenenbaum and Pollard's ODE book. The question is about determining whether functions are solutions to differential equations, and to state the interval where the differential equation makes sense. The part I'm stuck on is:
$$\text{(e) }xy'=2y, \text{ } y=x^2$$
The actual calculation to demonstrate that this is true is simple enough.
$$y'=2x\rightarrow xy'=2y\rightarrow 2x^2=2(x^2)\text{ }$$
However, the solution states that the interval where it makes sense is $x\neq 0$, which is confusing to me. I don't understand why the interval isn't $-\infty<x<\infty$, as neither the function nor the differential equation have any discontinuties at $x=0$. Am I missing something, or is this a mistake in the solutions? Any input would be greatly appreciated.
In any differential equation, the derivative must be well-defined for all values for which the equation is expected to hold. In your case, you have
$$xy' = 2y \tag{1}\label{eq1}$$
The singularity you mention comes from when you isolate $y'$ by itself to get
$$y' = \frac{2y}{x} \tag{2}\label{eq2}$$
As Aniruddha Deshmukh stated in a question comment, the right side is undefined at $x = 0$, so the derivative is not properly defined there and, thus, the solution is not either.