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!