I'm having some trouble finding the interval on which the solution of the following ODE is valid.
$$\begin{cases}y'(t)=\dfrac{1}{\cos(t)^2}\\y(9)=\tan(9)
\end{cases}$$
The solution is $y(t)=\tan(t)$.
On which interval is this solution valid? Thank you
Best Answer
Hint. The domain of differentiability of $\tan(t)$ where $$D(\tan(t))=\frac{1}{\cos^2(t)}$$ is $$\bigcup_{k\in\mathbb{Z}}\left(-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi\right).$$ Find the maximal interval in such domain which contains the initial value $t=9$.