# Trigonometric general solution to ordinary differential equation

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?

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Yes, it is as you wrote, the $n\pi$ is required. –  André Nicolas Nov 17 '12 at 19:10

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 $\left(-\frac{\pi}{2}+n\pi,\frac{\pi}{2}+n\pi\right)$.
Could you expand on the requirement of different constants $c_{n}$, I'm confused as to why they are necessary. Thanks! –  Shaktal Nov 17 '12 at 19:32
Look at the equation $\frac{dy}{dx}=\frac{1}{x}$, a simple (?) integration. It is standard, but not quite correct, to write $y=\ln|x|+C$. But you will notice that if we let $y=\ln(-x)+17$ when $x\lt 0$, and $y=\ln(x)+99$ when $x\gt 0$, then everywhere the dervative is defined, we have $\frac{dy}{dx}=\frac{1}{x}$. The issue is that the domain of definition of $\dfrac{1}{x}$ is not connected because of the singularity at $0$. The same issue arises in your problem: $\tan^2 y$ has singularities. However, in practice we will have an initial condition that removes all ambiguities. –  André Nicolas Nov 17 '12 at 19:39