fix(ch1): correct interval of validity, add chapter label

chapters/ch01_introduction.tex

@@ -1,5 +1,6 @@
 \section{Classification of Differential Equations}
 \label{sec:ch01_classification}
+\label{ch:introduction}
 
 A \textbf{differential equation} (DE) is a relation between an unknown function and its derivatives. Differential equations are the language of dynamical systems, appearing in physics, engineering, biology, economics, and throughout the applied sciences.
 
@@ -395,7 +396,11 @@ The interval of validity may be restricted by:
 
   Separating: $\displaystyle \frac{\diff y}{\sqrt{y}} = \diff x \;\Rightarrow\; 2\sqrt{y} = x + C$. With $y(0) = 1$, we get $C = 2$, so $y(x) = \left(\frac{x + 2}{2}\right)^2 = \frac{(x+2)^2}{4}$.
 
-  This solution is defined and differentiable for all $x \in \R$, so the interval of validity is $(-\infty, \infty)$. Note that although the Lipschitz condition fails at $y = 0$ (since $\pd{f}{y} = \frac{1}{2\sqrt{y}}$ is unbounded there), the particular solution passing through $y(0) = 1$ never reaches $y = 0$ for $x > -2$, so uniqueness is preserved along this trajectory.
+  To find the interval of validity, check that the solution satisfies the original ODE: $y' = \frac{x+2}{2}$ while $\sqrt{y} = \frac{|x+2|}{2}$. These are equal only when $x + 2 \geq 0$, i.e., $x \geq -2$. For $x < -2$, we have $y' < 0$ while $\sqrt{y} \geq 0$, so the formula fails the ODE. At $x = -2$, the solution reaches $y = 0$ (the equilibrium). Therefore, the interval of validity is
+  \[
+    (-2, \infty).
+  \]
+  Note that although the Lipschitz condition fails at $y = 0$ (since $\pd{f}{y} = \frac{1}{2\sqrt{y}}$ is unbounded there), the particular solution passing through $y(0) = 1$ is valid on $(-2, \infty)$, and uniqueness is preserved on this interval.
 \end{workedexample}
 
 \section{Summary}