Solve: $$\frac{dx}{dy}=(x^{2}-x-12)(1+\tan^{2}{y})$$
This is a first order, linear, separable ODE, so it can be solved by rearranging to:
$$\frac{dx}{x^{2}-x-12}=(1+\tan^{2}{y})\:dy$$
And then integrating both sides:
$$\frac{1}{7}\left[\int{\frac{dx}{x-4}}-\int\frac{dx}{x+3}\right]=\int\sec^{2}{y}\:dy \\ \therefore \frac{1}{7}\ln{\left|\frac{x-4}{x+3}\right|}=\tan{y}+c$$
Re-arranging to find $y$ as a function of $x$, we get:
$$y=\arctan{\left(\frac{1}{7}\ln{\left|\frac{x-4}{x+3}\right|}+c\right)}+n\pi,\quad n\in\mathbb{Z},c\in\mathbb{R}$$
My question is, is the additive $n\pi$ for arbitrary integer $n$ required in the solution. My assumption is that it should be required because $y$ represents a family of solutions $\forall c \in \mathbb{R}$, is this right?
Thanks in advance!
Yes, the $n\pi$ is required. One can see this already before going through the mechanics of solving the equation. The equation involves $\tan^2 y$ and $dy$, so cares only about the value of the trigonometric function, and whether it is growing or shrinking.
Indeed, the situation is somewhat more complicated than you indicate. For a truly general solution, we need possibly different constants $c_n$ over the intervals $\left(-\frac{\pi}{2}+n\pi,\frac{\pi}{2}+n\pi\right)$.