I'm wondering why this differential equation, $xy'= 1$ has no solution on the interval $I:−1 < x < 1$ per Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard, page 29: example 4.22.
From my understanding, this differential equation should have a solution as long as $x ≠ 0$.
It's "family" of solutions would then look like, $y = log|x| + c_1$ (which is given in the book), and seems to be defined for all $x ≠ 0$ as well.
What am I missing here?
It's not only that no solution exists for $x=0$. It's more that the singular point $x=0$ subdivides the solution domain, such that a solution can only exist in $(0,\infty)$ or $(-\infty,0)$. Any interval that contains $0$, such as $(-1,1)$, fails this requirement, so no solution can exist.
What about $y=\ln|x|+c$, you might ask. Well, this is only a shorthand notation for a general solution when no initial condition is specified. If an initial condition is given, e.g. $y(x_0)=y_0$, then only one of these solutions (but not both) can exist
\begin{align} y &= \ln(x) + c_1, && \text{if } x_0 > 0 \\ y &= \ln(-x) + c_2, && \text{if } x_0 < 0 \end{align}
For example, if $y(1)=0$ then the solution can only be $y=\ln(x)$, and if $y(-1)=0$ then the solution can only be $y=\ln(-x)$