I have the following initial values problem:
\begin{equation} \left\{\begin{array}{@{}l@{}} y'(t)= t\tan(y(t) - \frac{\pi}{4})\\ y(0) = \frac{\pi}{2} \end{array}\right.\,. \end{equation}
We firstly observe that the RHS of the ODE is a function $f(y, t)$ defined in $\mathbb{R} \times (\mathbb{R} -\{\frac{3\pi}{4} + k\pi\})$ satisfying the hypothesis of the Cauchy-Lipschitz theorem, thus the problem locally (in an interval $I$) admits a unique solution, denoted by $y$.
Applying separation of variables gives the equation $$ \log|(\sin(y(t)-\frac{\pi}{4}))| = \frac{t^2}{2} + c $$
From this result, my book states that given the constant solution of the ODE $z(t) = \frac{\pi}{4}$, and given that $y(0) = \frac{\pi}{2} > \frac{\pi}{4} = z(0)$, then $$ \frac{\pi}{4} < y(t) < \frac{3\pi}{4} $$
for every $t \in I$, and proceeds to cancel the absolute value, to express $y$ in a nicer explicit form.
I don't understand how the $\frac{\pi}{4} < y(t)$ follows. I know that, if $f$ satisfied the conditions of the $\textit{global}$ existence and uniqueness theorem, we could say that two different solutions of the ODE don't intersecate, but in this case the author is using just the fact that there exists a local solution. Maybe there is a proposition I don't know, or the book is using other properties of this particular problem?
Assume the inequality $\frac{\pi}{4}<y(t)$ is wrong. Using the intermediate value theorem, there must be a $t_1$ with $y(t_1)= \frac{\pi}{4}$. Look at the IVP $z'(t)=t\tan(z(t)-\frac{\pi}{4})$, $z(t_1)=\frac\pi4$, then there are two solutions: $y$ and the constant $\frac{\pi}{4}$ are solutions. By uniqueness of the IVP, $y\equiv \frac\pi4$. But $y(0)=y_0 =\frac\pi2>\frac\pi4.$