Here's an exercise from Tenebaum & Pollard's book:

I know how to show that the bottom one is the solution of the top equation, what I don't understand is why the interval on which the solution works is not:Differential equation: $\displaystyle x \frac{dy}{dx} = 2y$

Solution: $\displaystyle y = x^{2}$

Give the interval on which the solution makes sense.

$\displaystyle -\infty < x < \infty$

but instead:

$\displaystyle x \neq 0$ as given the by author.

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Trying to understand this intuitively, I'm thinking that anything multiplied by zero is zero. So when $\displaystyle x = 0$, anything substituted on the left hand side gives zero. However, if $\displaystyle y = x^{2}$ is the solution, when $\displaystyle x = 0$, the RHS is always zero.

I don't see where the restriction $\displaystyle x \neq 0$ comes from.

Thanks for your help!