2
$\begingroup$

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!

  • 0
    Yes, it is as you wrote, the $n\pi$ is required.2012-11-17

2 Answers 2

1

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 $\left(-\frac{\pi}{2}+n\pi,\frac{\pi}{2}+n\pi\right)$.

  • 0
    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.2012-11-17
0

Apparently the goal was to find all solutions. If one may pardon the use of such locutions, the arctangent is a "multiple-valued function", and someone made a point of including all of its values. I.e. the purpose at that step was to find all numbers whose tangent is a certain number.