Here is an initial value problem =
$$ \frac{{\rm d}y}{{\rm d}t} = \sec^2t $$
subject to the condition that $y(30) = \tan(30)$.
Question = Write $y$ as a function of $t$. Also, on what interval is the solution valid?
My attempt =
So $y$ is quite simply equal to $\tan t$. The real issue is the interval of validity. The general solution to this type of problem is to write the equation as $y'(t) + p(t) y = g(t)$ subject to the condition $y(t_n) = y_n$,
So in this case we know that $t_n = 30$ and $p(t)$ seems to be zero. We know $t_n$ must be contained inside the interval. The second thing we know is that all values for which $p(t)$ and $g(t)$ are discontinuous must be excluded form the interval. So basically all we need is an interval that (i) contains 30 and (ii) ensures that $\sec^2t$ is continuous. Thus my solution is that $26.84 < t < 30.0022$
But this answer is wrong. Can someone kindly explain? Thank you.
Seems to me that the entire issue is based on radians/degrees. Take $tan(30)=tan(\pi/6) = \sqrt{3}/3$
Unless I'm mistaken, $Sec^2(t)$ has vertical asymptotes at $t=\frac {n\pi}2$ where $n$ is any integer, thus your interval would be $[\frac{-\pi}2,\frac{\pi}2]$.